These Statistics NOTES Belong to:

These Statistics NOTES Belong to: Topic Notes Questions Date 1 2 3 4 5 6 REVIEW DO EVERY QUESTION IN YOUR PROVINCIAL EXAM BINDER Important Calculator Functions to know for this chapter Normal Distributions Calculator function What it finds. Key words to look for: Cdfnorm(leftmost#, rightmost#, mean, standard deviation) Invnorm(probability, mean, standard deviation) Binomial Distribution Probability (Area under the curve) Z-score or cut off value that cuts of leftmost area. Normal distribution, what is the probability? Normal distribution, what value cuts off the first 30% under the curve? Calculator function What it finds. Key words to look for PdfBin(# of choices, probability, exact # out of total) Probability of a single event Binomial distribution, Find the probability of exactly 9 out of 20 heads. CdfBin(# of choices, probability, up to how many out of total) Cumulative probability Binomial distribution, Find the probability of at most 9 out of 20 heads. Cdfnorm(leftmost#-0.5, rightmost#+0.5, mean, standard deviation) Probability of a single event of cumulative Normal approximation to the binomial. Math12principles/tspraynotes/2005/copyright tspray2005 1

Binomial Distribution Frequency Distribution Normal Distribution Probability Distribution Standard Normal Distribution Uniform Distribution What definitions and formulas should I know? A function that describes the outcomes of an experiment. A function where only 2 outcomes are possible! ie T/F or Heads/ Tails. A function often described using a table of values or a histogram, that provides the frequency for every outcome of an experiment. A probability function with mean µ and a standard deviation of σ ; the graph is symmetrical about the mean; abides by the 68-95-99.7 rule. A function that provides the probability for every outcome of an experiment. A probability function with mean 0 and a standard deviation of 1; the graph is symmetrical about the mean; obeys the 68-95-99 rule. A probability function where the probability for each event is equal. µ Pronounced Mew The mean. x Pronounced X-bar x i µ = n σ Pronounced Sigma σ = Mean Median Mode ( µ ) x i n 2 Standard Deviation The mean. x 1, x2, x3,... x n represent a population --> Remember means sum of n is size of the population Standard deviation. x 1, x2, x3,... x n represents a population µ represents the mean n represents the population size The sum of a set of numbers divided by the number of numbers in the set. The middle number when data is arranged in sequential order. The number that occurs most often in a set of numbers. A measure of the extent to which data cluster around the mean. Binomial Distribution A function where only 2 outcomes are possible! ie T/F or Heads/ Tails. For a binomial population the mean can be found by µ = np Mean µ = np n represents the population size p Probability of success Standard Deviation σ = np(1 p) σ = npq For a binomial population the standard deviation q is equal to 1-p which represents the probability of not success What is the 68-95-99 rule? 68-95-99 rule About 68% of the population are within 1 standard deviation of the mean; about 95% of the population are within 2 standard deviations of the mean; about 99.7% of the population are within 3 standard deviations of the mean What is a Z- Score? Z-score or Z-value µ Z = x σ A z-score is a standard deviation. More specifically, positive and negative standard deviations from the mean, 0, on the standard normal curve. Use this formation to turn values from the normal curve into z-scores. Math12principles/tspraynotes/2005/copyright tspray2005 2

What is a Distribution? Frequency Distribution: Is obtained by performing experiments. Experiments will vary. Probability Distributions: A probability distribution is a function that provides the probability for every outcome of an experiment. Uniform Distribution Binomial Distribution The probability of each event is equal. Example: Role a die. The uniform distribution is given below. Has a fixed number of independent and identical trials Each trial has 2 possible outcomes. (Yes or No, head or tail, pass or fail, win or lose, 2 or not a 2.) Example: Flip a coin 4 times The sample space is given below: HHHH, HHHT, HHTH, HTHH,THHH, HHTT, HTTH, TTHH, HTHT, THTH, THHT, HTTT, THTT, TTHT, HHHT, TTTT, Example: Spin a pointer. The uniform distribution is given below. The binomial distribution for the number of heads in 4 tosses of a coin is given below: The total area of the bars in each probability distribution adds to 1. Math12principles/tspraynotes/2005/copyright tspray2005 3

Write a question that would result in a uniform distribution. A spinner with 4 options 1. 2. Write a question that would result in a binomial distribution. A dice with 6 options 3. 4. Creating a Binomial Distribution: Flick a spinner 4 times Determine the binomial distribution of the number of times a S comes up as a table. Role a die 4 times Determine the binomial distribution of the number of times a 5 comes up as a table. Frequency Formula Probability Frequency Formula Probability 0 S s 1 P(0) = 4 C 0 0 3 4 4 5. 6. 4 = 1 S 2 S s 3 S s 4 S s 1 P(1) = 4 C 1 1 4 3 3 4 = 1 P(2) = 4 C 2 2 4 3 2 4 = 1 P(3) = 4 C 3 3 4 3 1 4 = 1 P(4) = 4 C 4 4 4 3 0 4 = A faster way using your graphing calculator: x 1 3 Press Y=, type in P(x) = 4 C x 4 4 7. 8. 9. 10. 11. 12. 13. 14. 4 x, Press Table, record numbers from the table. Math12principles/tspraynotes/2005/copyright tspray2005 4

CALCULATING MEAN AND STANDARD DEVIATION Mean Standard deviation Median Mode The mean is the sum of the measurements divided by the total number of measurements. The standard deviation is a measure of how spread out a distribution is. The median is the middle number in a set of ordered numbers. The mode is the event that occurs most often in a set of numbers. µ = σ = n X i i =1 n n i =1 (X i µ) 2 n 2,2,2,3,4,5,5,5,6 2 & 5 from the above data Frequency Frequency is the measure of how often an event occurs. The frequency of twos in the above data is 3. Determine the mean and standard deviation: 15. Find the mean of 4,5,6,7,8. 16. Find the standard deviation of the numbers, 4,5,6,7,8. µ = 4 + 5 + 6 + 7 + 8 5 µ = 6 Solution σ = σ = σ = 2 ( 4 6) 2 + ( 5 6) 2 + ( 6 6) 2 + ( 7 6) 2 + ( 8 6) 2 4 + 1 + 0 + 1 + 4 5 5 17. Find the mean of 17,8,18,24. 18. Find the standard deviation of the numbers 17,8,18 and 24. 19. Find the mean of 89,45,65. 20. Find the standard deviation of the numbers 89,45 and 65. 21. Find the mean of 50,60,68,72. 22. Find the standard deviation of the numbers, 24, 56 & 34. Math12principles/tspraynotes/2005/copyright tspray2005 5

Determine the mean and standard deviation for A > 0 & B > 0 : 23. Calculate the mean of A-B,A,A+B 24. Calculate the standard deviation of A-B,A,A+B µ = µ = 3A 3 µ = A n X i i =1 n = A B + A + A + B 3 25. Calculate the mean of A-3,A,A+6 σ = n i =1 (X i µ) 2 n σ = B2 + B 2 3 = = 2B2 3 = B 2 3 (A B A) 2 + (A A) 2 + (A + B A) 2 26. Calculate the standard deviation of A-3,A,A+6 3 A+1 27. Calculate the mean of 2A+2B,2A,2A-2B 28. Calculate the standard deviation of 2A+2B,2A,2A-2B 3.74 29. Calculate the mean of A-5B, A, A+5B 30. Calculate the standard deviation of A-5B, A, A+5B 31. Calculate the mean of A+B, 3A-2B, B+2A 32. Calculate the standard deviation of A+B, 3A-2B, B+2A OMIT 33. Calculate the mean of 2A+3B,2A-B,-2B-A 34. Calculate the standard deviation of 2A+3B,2A-B,-2B-A OMIT Math12principles/tspraynotes/2005/copyright tspray2005 6

Find the mean, median and mode of the following data. 2,3,4,4,4,5,6 1,1,3,4,5,6,8 0,1,2,3,4,4,14 0,0,0,0,0,4,24 0,0,1,4,15 Mean 35. 36. 37. 38. 39. Median 40. 41. 42. 43. 44. Mode 45. 46. 47. 48. 49. Determine the mean and standard deviation: 50. The length and frequency of arbutus trees for a particular area in Gordon Head is given. Determine the mean and standard deviation. 51. The score and frequency for a golfer is given below. Determine the mean and standard deviation for this set of data. Length 10 12 14 16 Score +4 +5 +7 +9 Frequency 2 4 8 6 Frequency 6 3 8 2 Find the mean. µ = µ = n X i i =1 n = 2 10 ( ) + 4 ( 12) + 8 ( 14) + 6 ( 16) 2 + 4 + 8 + 6 Find the standard deviation. σ = σ = n i =1 (X i µ) 2 n 2 ( 10 ) 2 + 4 ( 12 ) 2 + 8 ( 14 ) 2 + 6 ( 16 ) 2 2 + 4 + 8 + 6 52. The length and frequency of arbutus trees for a particular area in Oak Bay is given. Determine the mean and standard deviation. 53. The score and frequency for a golfer is given below. Determine the mean and standard deviation for this set of data. Length 10 12 14 16 18 Score +4 +5 +7 +9 Frequency 24 40 15 6 14 Frequency 12 31 18 21 Math12principles/tspraynotes/2005/copyright tspray2005 7

Solve for x. 54. 15 students take a test and their scores are listed below. Determine the value of x if the mean score is 30. 55. 20 students take a test and their scores are listed below. Determine the value of x if the mean score is 7.9. Score 40 50 20 x Score 10 20 60 x Frequency 5 3 2 5 Frequency 5 5 8 2 Fill out the mean formula and simplify 5(40) + 3(50) + 2(20) + 5(x) 30 = 15 200 + 150 + 40 + 5x 30 = 15 Cross multiply and solve for x 30 15 = 390 + 5x 450 = 390 + 5x 60 = 5x x = 12 56. 30 students take a test and their scores are listed below. Determine the value of x if the mean score is 10.8 OMit 57. 50 students take a test and their scores are listed below. Determine the value of x if the mean score is 19.2 Score 11 22 x 15 Score 14 x 56 15 Frequency 3 6 12 9 Frequency 5 15 10 20 Math12principles/tspraynotes/2005/copyright tspray2005 8

58. Determine the mean, median, mode & standard deviation using your graphing calculator 59. Determine the mean, median, mode & standard deviation using your graphing calculator Score 11 22 10 15 Score 14 12 56 15 Frequency 3 6 12 9 Frequency 5 15 10 20 Press Stat Edit fill out table Press 4 operations key( this gets you to the calculator setting. Press stat choose Calc choose 1_stats type L1 type, type L2, press enter X = mean σ X = σ 60. Determine the mean, median, mode & standard deviation using your graphing calculator 61. Determine the mean, median, mode & standard deviation using your graphing calculator Score 11 22 10 25 Score 14 12 56 15 Frequency 3 6 12 9 Frequency 5 15 10 20 OMIT 62. Determine the mean, median, mode & standard deviation using your graphing calculator 63. Determine the mean, median, mode & standard deviation using your graphing calculator Score 20 25 11 15 Score 4 2 56 105 Frequency 5 12 3 90 Frequency 51 15 12 20 Math12principles/tspraynotes/2005/copyright tspray2005 9

Standard deviation is the measure of how much the data varies. 64. Arrange the each distribution with mean, µ, in order from the greatest standard deviation to smallest standard deviation. A. B. C. D. Biggest standard deviation Smallest standard deviation 65. Each individual bar encloses the same area. Arrange the each distribution with mean, µ, in order from the greatest standard deviation to smallest standard deviation. A. B. C. D. E. F. Biggest standard deviation Smallest standard deviation Math12principles/tspraynotes/2005/copyright tspray2005 10

Formulas for the Mean and Standard Deviation for Binomial Distributions Mean: µ = n p Standard deviation σ = n p q Where n is the number of trials, p is the probability of success on each trial and q is the probability of failure on each trial. Find the mean and standard deviation for 66. The number of heads that occur when a coin is tossed 80 times. 67. The number of 5 s that occur when a die is rolled 120 times. 68. The number of even numbers that occur when a die is rolled 190 times. 69. The number of heads when a biased coin with P(H)=0.7 is tossed 90 times. 70. The number of heads that occur when a coin is tossed X 2 times. 71. The number of 5 s that occur when a die is rolled X 4 times. 72. The number of heads when a biased coin with P(H)=P is tossed 100 times. 73. The number of heads when a biased coin with P(H)=P 2 is tossed 80 times. 74. The number of heads that occur when a coin is tossed 270 times. 75. The number of 2 s that occur when a die is rolled 210 times. 76. The number of even numbers that occur when a die is rolled 790 times. 77. The number of heads when a biased coin with P(H)=0.7 is tossed 620 times. 78. A fair sided die is rolled 60 times. Calculate the mean of the binomial distribution for the number of times a 5 appears. 79. A fair sided die is rolled 70 times. Calculate the mean of the binomial distribution for the number of times a 3 appears. 80. A four sided die is rolled 896 times. Calculate the mean of the binomial distribution for the number of times a 2 appears. Math12principles/tspraynotes/2005/copyright tspray2005 11

81. A fair four-sided die is rolled n times. If the standard deviation of the number of times a 5 comes up is 12, determine the value of n. 82. A fair six-sided die is rolled n times. If the standard deviation of the number of times a 1 comes up is 20, determine the value of n. 83. A fair four-sided die is rolled n times. If the standard deviation of the number of times a 3 comes up is 18, determine the value of n. 84. A biased coin is tossed 100 times. The standard deviation for the number of heads that occurs is 3. If heads seems to be happening more than tails, determine the probability of heads. 85. A biased coin is tossed 2400 times. The standard deviation for the number of heads that occurs is 24. If heads seems to be happening more than tails, determine the probability of heads. 86. A biased coin is tossed 64 times. The standard deviation for the number of heads that occurs is 3.2. If heads seems to be happening more than tails, determine the probability of heads. σ = n p q 3 = 100 p (1 p) Square both sides 9 = 100p(1 p) Expand and solve for x 9 = 100p 100p 2 100p 2 100p + 9 = 0 Solve by factoring or use the quadratic formula. (Given Formula) x = b ± b2 4ac 2a x = ( 100) ± ( 100)2 4(100)(9) 2(100) x = x = x = 180 200 100 ± 6400 200 100 ± 80 200 = 0.9 or x = 20 200 = 0.1.6.8 87. A biased coin is tossed 96 times. The standard deviation for the number of heads that occurs is 4.8, If heads seems to be happening more than tails, determine the probability of heads. 88. A biased coin is tossed 252 times. The standard deviation for the number of heads that occurs is 4.762. If heads seems to be happening more than tails, determine the probability of heads. Since heads seems to be happening more often we will assume that P(H)=0.9 and P(T)=0.1.,.6.9 Math12principles/tspraynotes/2005/copyright tspray2005 12

89. A biased six-sided die is rolled 80 times. If the mean of the number of times a 5 comes up is 30, determine the value of p. 90. A biased six-sided die is rolled 120 times. If the mean of the number of times a 4 comes up is 90, determine the value of p. 91. A biased six-sided die is rolled 120 times. If the mean of the number of times a 3 comes up is 30, determine the value of p. 92. A biased six-sided die is rolled 200 times. If the mean of the number of times a 5 comes up is 30, determine the value of q. 93. A biased six-sided die is rolled 180 times. If the mean of the number of times a 4 comes up is 80, determine the value of q. 94. A biased six-sided die is rolled 150 times. If the mean of the number of times a 3 comes up is 60, determine the value of q. Properties of a Normal Distribution The Normal Curve Characteristics: Smooth, bell-shaped curve. Symmetric about the mean, µ. Asymptotic tails (taper off but never touch). Marked in standard deviations, ±σ on both sides of the mean. The median, mean, mode are the same value. The area under the curve is equal to one or 100%. 68% of the population is within 1 standard deviation of the mean. 95% of the population is within 2 standard deviations of the mean. 99% of the population is within 3 standard deviations of the mean. Math12principles/tspraynotes/2005/copyright tspray2005 13

Find the missing percentages. Determine the percentage of the population that is represented by each section under the normal curve. 95. 96. 97. 98. 99. 100. Use the 68-95-99 Rule to solve the following problems. The scores on a test are normally distributed with a mean score of 300 and a standard deviation of 50. Determine the percentage of scores that lie between 200 and 400 inclusive. 101. A score is between 200 and 400? 102. If 1000 students wrote this test, how many had scores between 200 and 400? 103. A score less than 200? 104. A score between 300 and 450? 105. A score higher than 350? 106. A score less than 400? 107. A score between 200 and 300? 108. A score between 150 and 450? 109. A score between 200 and 450? 110. A score between 200 and 400? The scores on a test are normally distributed with a mean score of 200 and a standard deviation of 25. Determine the percentage of scores that lie between 150 and 225 inclusive. 111. A score is between 150 and 225? 112. If 1000 students wrote this test, how many had scores between 150 and 225? 113. a score less than 175? 114. a score between 150 and 275? 115. a score higher than 250? 116. a score less than 275? Math12principles/tspraynotes/2005/copyright tspray2005 14

Statisticians use the letter Z to represent the standard normal curve. 68% of the data is within 1 standard deviation of the mean. Standard Normal Curve 95% of the data is within 2 standard deviations of the mean. Characteristics Special member of the normal curve family. µ=0 & σ=1. The increments on the horizontal axis of the standard normal curve all call z- scores. A z-score describes the number of standard deviations above or below the mean. 99% of the data is within 3 standard deviations of the mean. The standard normal curve can be used to compare normal distributions. Graphing Calculator Help 117. Find the shaded area. 118. Find the z-score. Finding the area finds the percentage & probability Calculator answers this question Area=P(-1<Z<2) CdfNorm(Left Z-score, Right Zscore, µ,σ) CdfNorm(-1,2,0,1)=0.81859 Statisticians use the letter Z to represent the standard normal curve. The area under the curve is equal to the probability that the z- scores are less than a. Find the 80 th percentile Calculator answers this question 0.8=P(Z<a) InverseNorm(Probability,µ,σ) InverseNorm(0.8,0,1)=0.84162 Statisticians call a a z-score. A z-score cuts off a portion of the area underneath the standard normal curve. Math12principles/tspraynotes/2005/copyright tspray2005 15

Find the percentage of the population is shaded. 119. P(Z < 1) = 120. P(Z > 2) = 121. P( 1 < Z < 2) = Cdfnormal(Left, Right,µ, σ) Cdfnormal(-10*, -1,0, 1) Cdfnormal(Left, Right, µ, σ) Cdfnormal(-2, 10*, 0, 1) Cdfnormal(Left, Right, µ, σ) Cdfnormal(-1, 2, 0, 1) Probability= 0.159 Percentage= 15.9% Probability= 0.977 Percentage= 97.7% Probability= 0.819 Percentage= 81.9% *99.7% of the population is between 3 and 3 standard deviations of the mean. Choosing a number smaller than 10 ensures that the lower portion of the population is enclosed. *99.7% of the population is between 3 and 3 standard deviations of the mean. Choosing a number larger than 10 ensures that the upper portion of the population is enclosed. If the mean and standard deviation are not included, cdfnormal will assume that the mean is 0 and the standard deviation is 1. Calculate the area of the shaded region under the standard normal curve as shown below. 122. P( 2 < Z < 1) = 123. P( 1 < Z < 0) = 124. P(Z < 2) = A variable, Z, is distributed as a standard normal. Determine each probability to 3 decimal places. 125. P(Z > 1) = 126. P( 2 < Z < 0) = 127. P(Z < 1) = A variable, Z, is distributed as a standard normal. Determine each probability to 3 decimal places. 128. Between -1.3 and 0.7 129. To the left of -1.8 130. Between -2.4 and -1.7 131. To the right of 1.3 Math12principles/tspraynotes/2005/copyright tspray2005 16

Find the Z-score given the area under the curve. 132. P(Z < a) = 0.8 133. P(Z > a) = 0.22 134. P(a < Z < 0) = 0.43 The calculator adds the area from left to right. It can only find the left most z-score if the area to found starts from the far right. To find the z-score for the rightmost 22%, calculate the zscore for 100%-22%=78%. To find the z-score for this portion of the curve, calculate the zscore for 50%-43%=7%. Z-score=InvNorm(Probability, µ,σ) Z-score=InvNorm( 0.8, 0,1)= 0.84 Z-score=InvNorm(Probability, µ, σ) Z-score=InvNorm( 1-0.22, 0, 1)= 0.77 Z-score=InvNorm(Probability, µ, σ) Z-score=InvNorm( 0.5-0.43, 0, 1)= -1.48 Determine the z-score. 135. P(Z < a) = 0.8 136. P(Z < a) = 0.62 137. P(Z < a) = 0.38 138. P(Z > a) = 0.38 139. P(Z > a) = 0.62 140. P(Z > a) = 0.80 141. P(a < Z < 0) = 0.38 142. P(0 < Z < a) = 0.28 143. P(0 < Z < a) = 0.48 144. 145. 146. Math12principles/tspraynotes/2005/copyright tspray2005 17

Draw and picture and explain the meaning of the following: 147. P(Z < a) = 0.9 148. P(Z > a) = 0.7 149. P( a < Z < a) = 80 Find the two values that cut off the following area under the standard normal curve. 150. 151. 152. Find the two z-scores that cut off 153. The central 60% 154. The central 35% 155. The central 20% 156. The central 30% Determine the area under the standard normal curve to 157. To the left of 0.3 158. To the right of 1.2 159. Between 1.4 and 2.7 160. Between -1.6 and 2.1 161. To the right of 2.3 162. To the left of 2 163. To the right of 2.1 164. To the left of -0.8 Math12principles/tspraynotes/2005/copyright tspray2005 18

Find the z-score. 165. P(Z < a) = 0.6 166. P(Z < a) = 0.8 167. P(0 < Z < a) = 0.28 168. P(0 < Z < a) = 0.4 169. P(Z > a) = 0.4 170. P(Z > a) = 0.46 171. P(a < Z < 0) = 0.37 172. P(a < Z < 0) = 0.17 173. P(0 < Z < a) = 0.24 174. P(0 < Z < a) = 0.2 175. P(Z < a) = 0.3 176. P(Z < a) = 0.56 177. P(a < Z < 0) = 0.24 178. P(a < Z < 0) = 0.41 179. P(Z > a) = 0.79 180. P(Z > a) = 0.26 181. For every normal distribution, about what percentage of the area under the curve is within 1 standard deviation of the mean? 182. Given a standard normal curve, determine the approximate value of P(-2<Z<2). 183. In a population that has a normal distribution with mean µ and standard deviation σ, determine the approximate percentage of the population that lies between µ -3σ and µ +3σ. Math12principles/tspraynotes/2005/copyright tspray2005 19

Modeling Real Situations Using Normal Distributions Z-score Conversion Formula z = x µ σ z is the z-score X is the particular data value µ is the mean σ is the standard deviation Given the following information, solve for the unknown. 184. Find the z-score: 75 68 z = 8 Subtract & simplify z = 7 8 z = 0.875 185. Find the standard deviation: 80 76 1.4 = σ Cross multiply & simplify 80 76 σ = 1.4 Simplify σ = 2.857 186. Find the mean: 1.5 = 146 µ 12 Cross multiply 1.5 12 = 146 µ Solve for µ 1.5 12 146 = µ µ = 164 Divide both sides by -1. µ = 164 187. Find x: 2.8 = x 20 7.6 Cross multiply 2.8 7.6 = x 20 Add 20 to both sides 2.8 7.6 + 20 = x Simplify x = 41.28 The average and standard deviation to 4 different tests are given. 188. Mean is 78 & Standard deviation 4 Find Timmy s z-score if he got 81 on the test. 189. Mean is 66 & Standard deviation 6 Find Timmy s z-score if he got 58 on the test. 190. Mean is 87 & Standard deviation 3 Find Timmy s z-score if he got 89 on the test. 191. Mean is 60 & Standard deviation 2 Find Timmy s z-score if he got 55 on the test. The average and standard deviation to 4 different tests are given. 192. Mean: 58 & Standard deviation 4 What did Timmy get on the test if his z-score is 2.4. 193. Mean: 66 & Standard deviation 6 What did Timmy get on the test if his z-score is 0.4. 194. Mean: 75 & Standard deviation 3 What did Timmy get on the test if his z-score is -1.4. 195. Mean: 60 & Standard deviation 2 What did Timmy get on the test if his z-score is -0.35. Math12principles/tspraynotes/2005/copyright tspray2005 20

The z-score and mean have been given. 196. z-score: 1.89 & mean: 79 197. z-score: -0.83 & mean: 67 198. z-score: -2.8 & mean: 81 199. z-score: 2.25& mean: 78 What is the standard deviation if Timmy got 82 on the test? What is the standard deviation if Timmy got 60 on the test? What is the standard deviation if Timmy got 48 on the test? What is the standard deviation if Timmy got 94 on the test? 200. The heights of a group of men are normally distributed with a mean height of 170 cm and a standard deviation of 9 cm. If the z-score for the height of one man is 1.8m, what is his height. 201. The heights of a group of men are normally distributed with a mean height of 190 cm and a standard deviation of 8 cm. If the z-score for the height of one man is 1.6, what is his height. 202. The scores on a test are normally distributed with a mean score of 300 and a standard deviation of 50. Determine the percentage of scores that lie between 180 and 346 inclusive. Determine: P(180 < x < 346) Solve using your graphing calculator: Cdfnormal(left, right, µ, σ)=probability Cdfnormal(180, 346, 300, 50)=0.813 Math12principles/tspraynotes/2005/copyright tspray2005 21

Onions at random have a mean mass of 180 grams and standard deviation of 20 grams. Assuming a normal distribution, what is the probability of choosing an onion with a mass between 100 grams and 220 grams? 203. A mass between 100 grams and 220 grams? 204. If Timmy has 80 onions, how many will weigh between 100 grams and 220 grams? 205. A mass between 100 grams and 200 grams? 206. A mass between 180 grams and 200 grams? 207. A mass between 190 grams and 250 grams? 208. A mass less than 200 grams? The volumes of mustard in bottles are normally distributed with a mean of 420 ml and a standard deviation of 3 ml. What proportion of the bottles contain less than 413mL? 209. Less than 413mL? 210. More than 419mL? 211. Find two values that cut off the central 40% 212. Find two values that cut off the central 60% 213. Find two values that cut off the central 80% The IQ scores for your class happen to be normally distributed with a mean of 110 and a standard deviation of 10. Determine the proportion that has an IQ less than 115. 214. Less than 115? 215. More than 84? 216. Between 88 and 112? 217. Between 98 and 112? 218. Between 98 and 122? 219. More than 126? 220. Find two values that cut off the central 50%. 221. Find two values that cut off the central 70%. 222. Find two values that cut off the central 10%. Math12principles/tspraynotes/2005/copyright tspray2005 22

Gigantic sheep at random have a mean mass of 200kgs and standard deviation of 60kgs. Assuming a normal distribution, what is the probability of choosing a gigantic sheep with a mass between 160kgs and 200kgs? 223. A mass between 100kgs and 200kgs? 224. If Timmy has 160 gigantic sheep, how many will weigh between 100kgs and 200kgs? 225. A mass between 110kgs and 250kgs? 226. A mass between 120kgs and 250kgs? 227. A mass between 200kgs and 250kgs? 228. A mass between 100kgs and 150kgs? 229. The mean and standard deviations are given for Biology are 63% and 18% and math is 69% and 5%. If Timmy has the same z-score in both biology and math, what is his math mark if his biology mark is 74%. 230. The mean and standard deviations are given for Biology are 60% and 9% and math is 79% and 9%. If Timmy has the same z-score in both biology and math, what is his math mark if his biology mark is 78%. 231. The mean and standard deviations are given for Biology are 70% and 20% and math is 70% and 4%. If Timmy has the same z-score in both biology and math, what is his math mark if his biology mark is 90%. Biology Z = x µ Z = σ 74 63 =0.611111 18 Math Timmy has the same z- score for math and biology. Z = x µ σ 0.611111 = x 69 5 x=72.055556 Math12principles/tspraynotes/2005/copyright tspray2005 23

Find the cut off values. 232. The results for a test are normally distributed. The mean score is 70 with a standard deviation of 9. If the top 10% of the students receive an A, what is the minimum mark needed to receive an A. 233. The results for a test are normally distributed. The mean score is 70 with a standard deviation of 10. If the central 30% get a C+, determine the upper and lower cut off marks. Calculators add the area under the curve from left to right. To find the z-score for the right-most 10%, calculate the z-score for 100%-10%=90%. Z-score=InvNorm(Probability, µ, σ) Z-score=InvNorm( 1-0.10, 70, 9) A mean of 70 means that 50% scored lower and 50% scored higher. Left z-score InvNorm(0.35*,70,10)= Right z-score InvNorm(0.65**,70,10)= *Left most z-score cuts off the first 35% of the population. 50%-15%=35% Since 50% are less than the mean and half of the cental 30% is 15%. **Right most z-score cuts off the first 65% of the popuation. 50%+15%=65% 234. A population of scores is normally distributed with a mean of 52.4 and a standard deviation of 14.3. If 30% of the scores are higher than a particular score x, calculate the value of x. 235. The diameters of oranges are normally distributed with a mean diameter of 8.5cm and a standard deviation of 0.8 cm. What is the largest diameter that would be less than 88% of the oranges? 236. A population of scores is normally distributed with a mean of 67 and a standard deviation of 12. If 30% of the scores are lower than a particular score x, calculate the value of x. 237. The diameters of oranges are normally distributed with a mean diameter of 9.5cm and a standard deviation of 0.7 cm. What is the largest diameter that would be less than 75% of the oranges? 238. A population of scores is normally distributed with a mean of 65 and a standard deviation of 7. If 80% of the scores are lower than a particular score x, calculate the value of x. 239. The diameters of oranges are normally distributed with a mean diameter of 8.5cm and a standard deviation of 0.8 cm. What diameter cuts off the largest 27.3% of the oranges? Math12principles/tspraynotes/2005/copyright tspray2005 24

240. Weight losses from a particular diet are normally distributed with a standard deviation of 1.2 k.g. If 58% of the weight losses are 6.5 kg or more, determine the mean weight loss. 241. Weight losses from a particular diet are normally distributed with a standard deviation of 3.2kgs. If 60% of the weight losses are 8.5 kg or more, determine the mean weight loss. 242. Weight losses from a particular diet are normally distributed with a standard deviation of 3.5kgs. If 48% of the weight losses are 2.5 kg or less, determine the mean weight loss. Write down what you know: Find the mean using: z = x µ z = 6.5 µ σ 1.2 You need to find the z-score before you can find the mean. 243. Weight losses from a particular diet are normally distributed with a standard deviation of 5kgs. If 60% of the weight losses are 12kgs or more, determine the mean weight loss. 244. Weight losses from a particular diet are normally distributed with a standard deviation of 3.5kgs. If 58% of the weight losses are 12.5kgs or less, determine the mean weight loss. InvNorm(0.42*,0,1)=-0.2 *100%-58%=42% z = x µ 0.2 = 6.5 µ σ 1.2 0.2 1.2 = 6.5 µ 0.2 1.2 6.5 = µ µ = 6.74 µ = 6.74 Math12principles/tspraynotes/2005/copyright tspray2005 25

The Normal Approximation to the Binomial Distribution 245. How many heads are possible for 6 flips? 246. How many heads are possible for 14 flips? 247. How many heads are possible for 32 flips? As the number of trials(flips) increases, the distribution looks more and more like a normal distribution. If the number of trials is big enough we can use the binomial distribution/function to solve normal approximation questions. The larger the number of trials the more accurate the approximation becomes. Graphing Calculator Help NormalCDF calculator function calculates the area under the curve from left to right. Remember if we wanted to use NormalCDF that we need the following: Normalcdf(Left,Right,µ,σ). Suppose we wanted to calculate the probability of 4 heads out of 14 flips. Typing in 4 will not work because 4 will not create any area under the curve. Typing in 4 will just leave a vertical line. To solve a binomial question using the normal approximation to the binomial we need to create area for exactly 4 heads. We can do this by typing in 3.5, 4.5 which will create the desired area. A fair coin is tossed 14 times. Determine the probabilities for the following number of heads using the normal approximation of the binomial. 248. Exactly 4 heads. 249. Between 3 and 8 heads inclusively. 250. More than 9 heads. Normalcdf(Left,Right,µ,σ) Normalcdf(3.5,4.5,np, npq ) N=14, p=0.5, q=0.5 Normalcdf(3.5,4.5,7, npq ) Math12principles/tspraynotes/2005/copyright tspray2005 26

A fair coin is tossed 14 times. Determine the probabilities for the following number of heads using the normal approximation of the binomial. 251. 252. 253. 254. 255. 256. 257. Use the normal approximation to the binomial distribution to calculate the probability of correctly guessing between 10 to 14 answers inclusive on a 25 question T-F test? 258. Use the normal approximation to the binomial distribution to calculate the probability of less than 14 5 s when a die is rolled 20 times. 259. Use the normal approximation to the binomial distribution to calculate the probability of correctly guessing less than 12 answers right on a 25 question T-F test? 260. Use the normal approximation to the binomial distribution to calculate the probability of more than 18 5 s when a die is rolled 20 times. 261. Use the normal approximation to the binomial distribution to calculate the probability of correctly guessing more than 17 answers right on a 25 question T-F test? 262. Use the normal approximation to the binomial distribution to calculate the probability of between 5-11 3 s (inclusive) when a die is rolled 20 times. Math12principles/tspraynotes/2005/copyright tspray2005 27

Can the normal approximation to the Binomial always be used to solve binomial distribution questions? NO Number of trials of n must be fairly large. Binomial distribution that is being approximated is fairly symmetric. The word fairly is satisfied when both np > 5 andnq > 5. Determine the following probabilities using the normal approximation to the binomial. State whether it is appropriate or not to use the normal approximation to the binomial for each question. 263. Situation: 264. Situation: 265. Situation: Coin is flipped 90 times P(H)=0.8. Coin is flipped 80 times P(H)=0.95. Coin is flipped 8800 times P(H)=0.98. Determine the probability that between 70 and 74 (inclusive) heads will occur? Determine the probability that between 50 and 76 (inclusive) heads will occur? Determine the probability that between 200 and 4900 (inclusive) heads will occur? 266. Situation: 267. Situation: 268. Situation: A fair coin is flipped 8 times. A fair coin is flipped 16 times. A fair coin is flipped 800 times. Determine the probability that between 4 and 5 (inclusive) heads will occur? Determine the probability that between 8 and 12 (inclusive) heads will occur? Determine the probability that between 200 and 400 (inclusive) heads will occur? Math12principles/tspraynotes/2005/copyright tspray2005 28

A multiple-choice test has 30 questions. Each question has 5 choices, only one of which is correct. If a student answers all the questions by randomly guess in, determine the probability that the student will correctly answer between 11 and 13 questions inclusive by using the following methods. 269. Use the binomial distribution to obtain this probability. 270. Use the normal approximation to obtain this probability. 74% of Math 12 students also take Physics. If 68 math 12 students are randomly selected, determine the probability that exactly 48 of these students also take physics course by using the following methods. 271. Use the binomial distribution to obtain this probability 272. Use the normal approximation to the binomial to obtain an estimate of this probability. A multiple-choice test has 60 questions. Each question has 4 choices, only one of which is correct. If a student answers all the questions by randomly guess in, determine the probability that the student will correctly answer between 14 and 15 questions inclusive by using the following methods. 273. Use the binomial distribution to obtain this probability 274. Use the normal approximation to obtain this probability. Math12principles/tspraynotes/2005/copyright tspray2005 29

64% of English 12 students also take English Literature. If 40 English 12 students are randomly selected, determine the probability that exactly 25 of these students also take English Literature course by using the following methods. 275. Use the binomial distribution to obtain this probability 276. Use the normal approximation to the binomial to obtain an estimate of this probability. A multiple-choice test has 40 questions. Each question has 4 choices, only one of which is correct. If a student answers all the questions by randomly guessing, determine the probability that the student will correctly answer between 15 and 17 questions inclusive by using the following methods. 277. Use the binomial distribution to obtain this probability. 278. Use the binomial distribution to obtain this probability. 54% of Math 12 students also take Physics. If 100 math 12 students are randomly selected, determine the probability that exactly 54 of these students also take physics course by using the following methods. 279. Use the binomial distribution to obtain this probability. 280. Use the normal approximation to the binomial to obtain an estimate of this probability. Math12principles/tspraynotes/2005/copyright tspray2005 30

The life expectancy of a car tire produced by a particular plant is normally distributed with a mean of 50 000 km and a standard deviation of 3000km. 281. What percent of these tires lasts between 49 000 km and 52 000 km? 282. If the plant makes 80 000 tires, how many tires would be expected to last more than 55 000 km? The life expectancy of a car tire produced by a particular plant is normally distributed with a mean of 55 000 km and a standard deviation of 5000km. 283. What percent of these tires lasts between 44 000 km and 51 000 km? 284. If the plant makes 90 000 tires, how many tires would be expected to last more than 58 000 km? The life expectancy of a car tire produced by a particular plant is normally distributed with a mean of 52 000 km and a standard deviation of 3500km. 285. What percent of these tires lasts between 51 000 km and 58 000 km? 286. If the plant makes 50 000 tires, how many tires would be expected to last more than 50 000 km? Math12principles/tspraynotes/2005/copyright tspray2005 31

A fair die is rolled 750 times. 287. What is the mean number of (3 s) that occur in 750 rolls of a fair die? 288. Use the normal approximation to the binomial to estimate the probability of obtaining a (3) between 118 and 134 times inclusive. 289. What is the standard deviation of the number of (3 s) that occur in 750 rolls of a fair die? A fair die is rolled 950 times. 290. What is the mean number of (3 s) that occur in 950 rolls of a fair die? 291. Use the normal approximation to the binomial to estimate the probability of obtaining a (3) between 100 and 150 times inclusive. 292. What is the standard deviation of the number of (3 s) that occur in 950 rolls of a fair die? Math12principles/tspraynotes/2005/copyright tspray2005 32

Stats Answer Key 1. Answers may vary. Example: A pointer is spun 100 times. List all the possible probabilities. 2. Answers may vary. Example: A pointer is spun 100 times. List the probabilities for the number of S s in 100 spins. 3. Answers may vary. Example: A die is rolled 120 times. List all the possible probabilities. 4. Answers may vary. Example: A die is rolled 120 times. List the probabilities for the number of 3 s in 120 rolls. 5. 0.316 6. 0.335 7. 0.422 8. 0.402 9. 0.211 10. 0.201 11. 0.047 12. 0.054 13. 0.004 14. 0.008 15. 6 16. 1.41 17. 16.75 18. 5.72 19. 66.3 20. 17.99 21. 62.5 22. 13.37 23. A 24. 0.816 25. A+1 26. 2.449 27. 2A 28. 1.633B 29. A 30. 4.08B 31. 2A 32. OMIT 33. A 34. OMIT 35. 4 36. 4 37. 4 38. 4 39. 4 40. 4 41. 4 42. 3 43. 0 44. 1 45. 4 46. 1 47. 4 48. 0 49. 0 50. 13.8,1.89 51. 5.95,1.67 52. 12.91,2.61 53. 6.32,1.84 54. 12 55. -236 56. 3.667 57. 2****??? 58. 14,13,10,4.517 59. 15,22.2,15,16.95 60. 13,14,10,4.517 61. 15,16.57,15???? 62. 16.21,15,15,3.32 63. 4,30.67,4 64. d,a,b,c 65. a,e,c,d,b,f 66. 40,4.5 67. 20,4.1 68. 95,6.9 69. 63,4.3 70. 0.5x 2,0.5x 71. 1 6 x 4,0.373x 2 72. 100p, 10 p(p 1) 73. p 2 80, 4p 5(1 p 2 ) 74. 135,8.2,5.8 75. 35,5.4 76. 395,14.1 77. 434,11.4 78. 10 79. 11.7 80. 149.3 81. 768 82. 2880 83. 1728 84. 0.9 85. 0.6??? 86. 0.8??? 87. 0.6??? 88. 0.9??? 89. 0.375 90. 0.75 91. 0.25 92. 0.15 93. 0.444 94. 0.4 95. 2.35%??? 96. 13.5% 97. 34% 98. 34% 99. 13.5% 100. 2.35%??? 101. 95% 102. 950 103. 2.5% 104. 49.85% 105. 16% 106. 97.5% 107. 47.5% 108. 99.7% 109. 97.35% 110. 95% 111. 81.5% 112. 815??? 113. 16% 114. 97.35% 115. 2.5% 116. 99.85% 117. 0.819 118. 0.842 119. 0.159,15.9% 120. 0.977,99.7% 121. 0.819,81.9 122. 0.819 123. 0.341 124. 0.977 125. 0.159 126. 0.477 127. 0.841 128. 0.661 129. 0.036 130. 0.0.36 131. 0.903 132. 0.84 133. 0.77 134. 1.48 135. 0.84 136. 0.31 137. 0.31 138. 0.31 139. 0.31 Math12principles/tspraynotes/2005/copyright tspray2005 33

140. 0.84 141. 1.17 142. 0.77 143. 2.05 144. 1.88 145. 1.23 146. 0.74 147. 90% of the population is less than a. a is the 90 th percentile. 148. 70% of the population is greater than a. a is the 30 th percentile 149. 80% of the data is enclosed by a zscores. 150. ± 0.47 151. ± 1.28 152. ± 1.08 153. ± 0.84 154. ± 0.45 155. ± 0.25 156. ± 0.39 157. 0.618 158. 0.115 159. 0.077 160. 0.927 161. 0.011 162. 0.977 163. 0.018 164. 0.212 165. 0.25 166. 0.84 167. 0.77 168. 1.28 169. 0.25 170. 0.1 171. 1.13 172. -0.44 173. 0.64 174. 0.52 175. 0.52 176. 0.15 177. 0.64 178. 1.34 179. 0.81 180. 0.64 181. 68% 182. 95% 183. 99.7% 184. 0.875 185. 2.875 186. 164 187. 41.28 188. 0.75 189. 1.33 190. 0.67 191. 2.50 192. 67.60 193. 68.4 194. 70.8 195. 59.3 196. 1.59 197. 8.43 198. 11.79 199. 7.11 200. 186.2 201. 202.8 202. 0.813 203. 0.977 204. 78.18 205. 0.841 206. 0.341 207. 0.308 208. 0.841 209. 0.01 210. 0.63 211. 418.43-421.57 212. 417.48-422.52 213. 416.16-423.84 214. 0.691 215. 0.995 216. 0.565 217. 0.464 218. 0.7698 219. 0.0547 220. 103.26-116.74 221. 99.64-120.36 222. 108.74-111.26 223. 0.45 224. 72.32 225. 0.73 226. 0.71 227. 0.3 228. 0.15 229. 72.1? 230. 97 231. 74 232. 81.5 233. 66.1,73.9 234. 59.9 235. 9.44 236. 60.71 237. 9.97 238. 70.89 239. 8.99 240. 6.74 241. 9.3 242. 2.68??? 243. 13.25??? 244. 11.79??? 245. 246. 247. 248. 0.06 249. 0.78 250. 0.21 251. 0.09 252. 0.02 253. 0.90 254. 0.12 255. 0.91 256. 0.97 257. 0.67 258. 1 259. 0.5 260. 0 261. 0.05 262. 0.24 263. 0.49 264. Not appropriate to use since nq<5. 265. 0 266. Not appropriate to 267. 0.59 268. 0.52 269. 0.0247 use since np<5, nq<5. 270. 0.0197 271. 0.0865 272. 0.0896 273. 0.2339 274. 0.2319 275. 0.1269 276. 0.128 277. 0.0498 278. 0.0471 279. 0.0798 280. 0.0799 281. 37.81% 282. 3823.27 3823 complete tires. 283. 19.795% 284. 24682 complete tires 285. 56.92% 286. 35807.3 tires 287. 125 288. 0.5928 289. 10.21 290. 158.33 291. 0.2476 292. 11.4867 293. 294. Math12principles/tspraynotes/2005/copyright tspray2005 34

A variable, Z, is distributed as a standard normal. Determine the percentage of the population that is shades to 3 decimal places. Find the z-score or z-scores that cut off each percentage of the standard normal distribution. Math12principles/tspraynotes/2005/copyright tspray2005 35

Math12principles/tspraynotes/2005/copyright tspray2005 36

Good news: This booklet covers the last 10 provincial exams. Not one mistake in the answer key! Look over???? in answer key Things that will eventually be added: Definitions page Next term Answers key will be added by Tuesday or Thursday( mistakes are on their way) Text book Questions Tuesday or Thursday Formatting Next term Mixing it up review next term This booklet covers the last 10 provincial exams. Not bad for 20-25 hours! Math12principles/tspraynotes/2005/copyright tspray2005 37