( Business Mathematics 1) Weeks 11 & 12 Dr. Marco A. Roque Department of Mathematics Texas A&M University
Distribution of Random Variables In many situations is good to assign numerical values to the outcomes of an experiment. We can do that introducing the following concept. Definition A random variables is a rule that assigns a number to each outcome of an experiment. In other words, a random variable is any function from the sample space to the set of real numbers. Let us see some examples to illustarte this idea. Examples All the information about the experiment can be expressed using a table Outcome HHH 3 HHT 2 HTH 2 THH 2 HTT 1 THT 1 TTH 1 TTT 0 Value of X 1) A coin is tossed three times. Let the random variable X denote the number of heads that occur in the three tosses.find the value assigned to each outcome of the experiment by the random variable X.. Thus, we see that the event X=2 is given by { HH, HTH, THH}
Distribution of Random Variables 2) A coin is tossed repeatedly until a head occurs. Let the random variable Y denote the number of coin tosses in the experiment. What are the values for Y? All the information about the experiment can be expressed using a table Outcome H 1 TH 2 Value of X TTH 3 TTTH 4.... Definition A random variable is called finite discrete if it assumes only finitely many values and it is said to be infinite discrete if it only takes on infitely many values. Finally random variables is called continuous if the values it may assume comprise an interval of real numbers. Now, if we are going to deal with random variables instead of the events themselves, then it will be necessary to construct a probabiltiy distribution associated to the random variable defined, but before doing that we need to specify the properties satisfied by the distribution. Tghus, if x1, x2,, xn are the values assumed by the random variable X with associated probabilities p1 = P(X=x1), p2 =P(X=x2),..., pn = P(X=xn) respectively, then the probability distribution of a random variable X satisifies: a. 0 pi 1 b. p1 + p2 + + pn = 1..
Distribution of Random Variables b. P(X 2 ) = P(X=2) + P(X=3) = 3/8 + 1/8 = 4/8 = 1/2 Examples 1) A coin is tossed three times. Let the random variable X denote the number of heads that occur in the three tosses.find the value assigned to each outcome of the experiment by the random variable X. a. Find the probability distribution of the random variable X. b. What is the probability of obtaining at least two heads in the three tosses of the coin? a. x P(X=x) 0 1/8 1 3/8 2 3/8 3 1/8 2) Let X denote the random variable that gives the sum of the faces that fall uppermost when two fair dice are rolled. a. Find the probability distribution of the random variable X. b. What is the probability that the sum of the faces that fall uppermost is less than or equal to 5?Between 8 and 10 inclusive? a. x P(X=x) x P(X=x) 2 1/36 8 5/36 3 2/36 9 4/36 4 3/36 10 3/36 5 4/36 11 2/36 6 5/36 12 1/36 7 6/36
Probabilty distributions and Statistics Distribution of Random Variables Examples b. P(X 5) = P(X=2) + P(X=3) + P(X=4) + P(X=5) = 1/36 + 2/36 + 3/36 + 4/36 = 10/36 P( 8 X 10 ) = P(X=8) + P(X=9) + P(X=10)= 12/36 b. What is the probability that the number of cars observed waiting in line in any two-minute interval between 3 pm and 5 pm on a Friday is less or equal to three? Between 2 and 4 inclusive? Greater than six? a.- Using the relative frequency interpretation of the probability and if X denote the random variable given by the number of cars observed waiting in line, then we have 3) The following data give the number of cars observed waiting in line at the beginning of 2-minute intervals between 3 pm and 5 pm on acertain Friday at the drive-in telller of the Westwood Savings Bank and the frequency of ocurrence. Cars 0 1 2 3 4 5 6 7 8 Frequency of 2 9 16 12 8 6 4 2 1 ocurrence a. Find the probability distribution of the random variable X, where X denotes the number of cars observed waiting in line X = x 0 1 2 3 4 5 6 7 8 Probability.03.15.27.20.13.10.07.03.02 b. P(X 3) = P(X= 0) + P(X=1) + P(X=2) + P(X=3) =.03 +.15 +.27 +.20 =.65 P(2 X 4) = P(X=2) + P(X=3) + P(X=4) =.27 +.20 +.13 =.60 P( X > 6) = P(X= 7) + P(X= 8) =.03 +.02 =.05
Distribution of Random Variables Histograms A probability distribution of a randon variable may be exhibited graphycally by means of Histogram. To construct a histogram of a particular probability distribution, first locate the values of random variable on a number line. Then, above each such number, erect a rectangle with width 1 and height equal to the probability associated with the value of the random variable. Thus, the area of each rectangle is equal to the probability of the random variable associated to the experiment and the sum of all areas is equal to 1.
Probabilty distributions and Statistics Expected value An important parameter to evaluate in the statistical analysis of a data set is the concept of expected value.to this end, let us inroduce the basic idea of average or mean. Cars 0 1 2 3 4 5 6 7 8 Frequency of 2 9 16 12 8 6 4 2 1 ocurrence Find the average of the cars waiting in line. Definition The average, or mean, of the numbers x 1, x 2, x 3,, x n x= 2+9+16+12+8+6+4+2+1 3. 1 60 We can rewrite the above calculations as Is given by x= x 1 +x 2 +x 3 +... +x n n 0(2/60) + 1(9/60) + 2(16/60) + 3(12/60) + 4(8/60) + 5(6/60) + 6(4/60) + 7(2/60) + 8(1/60) Example The following data give the number of cars observed waiting in line at the beginning of 2-minute intervals between 3 pm and 5 pm on acertain Friday at the drive-in telller of the Westwood Savings Bank and the frequency of ocurrence. Each term contains two elements, the first one is the value of the random variable and the second one is the probability associated to it, and this the way we can give the next step through the defintion of the expected value.
x P(X=x) x P(X=x) Definition Let X denote the random variable that assumes the values x 1, x 2, x 3,, x n with associated probabilities p 1, p 2, p 3,, p n respectively. Then, the expected value 2 1/36 8 5/36 3 2/36 9 4/36 4 3/36 10 3/36 5 4/36 11 2/36 6 5/36 12 1/36 E ( x )=x 1 p 1 +x 2 p 2 +x 3 p 3 +...+x n p n 7 6/36. Example 1) Let X denote the random variable that gives the sum of the faces that fall uppermost when two fair dice are rolled. Find the expected value, E(X), of X. The probability distribution od the random variable variable X is given by Therefore the expected values is E(x)= 2(1/36) + 3(2/36) + 4(3/36) + 5(4/36) + 6(5/36) + 7(6/36) + 8(5/36) + 9(4/36) + 10(3/36) +11(2/36) + 12(1/36) = 7 As we were expecting!!!
Probabilty distributions and Statistics Using the table we found 2) The Island Club is holding a fund-raising raffle. Ten thousand tickets have been sold for $2 each. There will be a prize of $3000, 3 second prizes of $1000 each, 5 third prizes of $500 each, and 20 consolation prizes of $100 each. Let X denote the net winnings ( that is, winning less the cost of the ticket ) associated with a ticket, find E(X). Interpret your results. The values of the random variable X are ( 0 2) = -2, ( 100 2) = 98, ( 500 2) = 498, ( 1000 2) = 998, ( 3000 2) = 2998 and the probability distribution is x P(X=x) -2.9971 98.0020 498.0005 998.0003 E(X) = (-2)(.9971) + (98)(.0020) + (498)(.0005) + (998)(.0003) + (2998)(.0001) = - 0.95 The above result means that in the long run a participant will loss 95 cents per dollar on each raffle. 3) Mike and play a card game with a standard deck of 52 playing cards. Mikes selects a card from a well-shuffled deck and receives A dollars from Bill if the card selected is a diamond; otherwise Mike pays a Bill a dollar. Determine the value of A if the game is to be fair, that is, the expected value has to be equal to zero. Let X denote the random variable associated with mike's winnings. Then X takes on the value A with probability P(X=A) = ¼ and takes on the value of -1 with probaility P(X=-1) = ¾ if Mike loss. The expected values has to be equal to zero Therefore A = 3. E(X) = A( ¼ ) + (-1)( ¾ ) = 0 2998.0001
Odds In the everyday world the probability of occurrence of an event is given through the concept of odds in favor of or odds against. Definition If P(E) is the probability of an event E occurring, then 1. The oddss in favor of E occurring are OBS If the odds in favor are expressed as a / b, we say that the odds in favor are a to b. If the odds against are expressed as b / a, we say that the odds against are b to a. Examples 1) Find the odds in favor in winning a bet on red in American roulette. P ( E ) 1 P ( E ) = P ( E ) P (E c ) ; [P ( E ) 1] The probability that the roullet lands on red is 18 / 38, therefore the odds in favor of winning on a red bet are 2. The odds against E occurring are 18/38 1 18/38 = 18/38 20/38 = 9 10 1 P ( E ) P ( E ) = P (E c ) P ( E ) ; [P ( E ) 0] Now if the odss in favor of an event E ocurring are a / b then P ( E ) = a a+b
Example Consider the following situations a. The odds that San Francisco will win the World Series thys season are 7 to 5. b. The odds that it will not rain Tomorrow are 3 to 2. Express each of these odds as a probability of the event ocurring. a. P ( E ) = a a+b = 7 5+7 = 5 12 Median and Mode Another two measures of central tendency are the median and mode defined as follows. Definition The median of a group of numbers arranged in increasing or decreasing order is: (a)the middle number if there is an odd number of entries. (b) the mean of the middle numbers if there is an even number of entries Example a. The times, in minutes, Susan took to go to work on nine consecutive working days were: 46 42 49 40 52 48 45 43 50 b. P ( E ) = a a+b = 3 3+2 = 3 5 b. The times, in minutes, Susan took to return home from work on eight consecutive working days were: 37 36 39 37 34 38 41 40 What is the median of her evening commute times?
a. Arranging the numbers in increasing order, we have 40 42 43 45 46 48 49 50 52 therefore the median is 46. b. Arranging the numbers in increasing order, we have 34 36 37 37 38 39 40 41 a. there is no mode. b. the mode is 3 c. there are two modes, namely, 3 and 4. therefore the median is ( 37 + 38 ) / 2 = 37.5 Definition The mode of a group of numbers is the number in the group that occurs the most. Example Find the mode, if there is one, of the gicen group of numbers. a. 1,2,3,4,6 b. 2,3,3,4,6,8 c. 2,3,3,3,4,4,4,8
Variance and Standard Deviation To this point we have discussed three concepts mean, median, and mode which represent the Measures of Central Tendency, but now we have to introduce a new concept which measures the way the data are concetrated or not around the mean value. Thus, for instance if we are looking for the height of students in High Schools, we can come up for two differet high schools having the same mean but that information is not enough to say how the data is spread out about the mean value. In this way, we have to introduce a new concept. In fact, we will introduce two concepts assocciated to this idea, known as Measures of Spread. Thus, the variance is giving to us a measure of how the dat set move apart from the mean value. Now, since the variance is considering the square of the deviation from the mean, then we need to introcuce another concept. Definition The Standard Deviation of a random variable denoted by σ ( sigma ), is defined by σ = ( x 1 μ) 2 p 1 +(x 2 μ) 2 p 2 +(x 3 μ ) 2 p 3 +...+(x n μ) 2 p n Definition Let X denote the random variable that assumes the values x 1, x 2, x 3,, x n with associated probabilities p 1, p 2, p 3,, p n respectively, and expected value E(X) = μ. Then, the variance of the random vraiable is Let us see some examples to illustrate the use of these concepts. Var ( X )=(x 1 μ) 2 p 1 +(x 2 μ ) 2 p 2 + (x 3 μ) 2 p 3 +...+(x n μ) 2 p n
Variance and Standard Deviation Examples 1) Find the variance of the random variable X and the random variable Y whose probability distributions are shown below. μ X =(1 ) (0.05 )+(2 ) (. 075 )+(3 ) (.20 )+...+(7 ) (. 05 ) = 4 μ Y =(1 ) (0. 2)+(2) (.15 )+ (3 ) (.1 )+...+(7) (.25 ) = 4 X P(X=x) Y P(Y=y) 1.05 1.2 2.075 2.15 3.20 3.1 4.375 4.15 5..15 5.05 6.10 6.1 7. 05 7.25 and the variance of each random variable is given by Var ( X )=(1 4 ) 2 (0.1 )+(2 4 ) 2 (.075 )+(3 4 ) 2 (.2)+...+(7 4 ) (.05 ) Var ( X ) = 1. 95 Var (Y )=(1 4 ) 2 (0.2 )+ (2 4) 2 (. 15 )+(3 4 ) 2 (.1)+...+(7 4 ) (. 25 ) Var ( X ) = 5. 2 In this way, we can can see that data associated to X are more concetrated to the mean than the values associated to the random variable Y. The mean of each random variable is given by
Variance and Standard Deviation Examples 2) let X and Y denote the random variables whose are the weights of the Brand A and Brand B potato chips, respectively. Compute the mean and standard deviations of X and Y, and interpret your results μ X =(15.8 ) (0. 1)+(15. 9 ) (..2 )+...+(16. 2 ) (.1 ) = 16 μ Y =(15.7 ) (0.2 )+(15.8 ) (. 1 )+ (15. 9 ) (.1 )+...+ (16.3 ) (.1 ) = 16 and the standard deviations are given by X P(X=x) Y P(Y=y) 15.8.1 15.7.2 15.9.2 15.8.1 16.0.4 15.9.1 16.1.2 16.0.1 16.2.1 16.1.2 16.2.2 σ = (15.8 16 ) 2 (.1 )+ (15.9 16) 2 (.2)+...+(16. 2 16 ) 2 (.1) σ = 0.11 σ = (15.7 16) 2 (. 2)+(15. 8 16) 2 (. 1)+...+(16.3 16 ) 2 (. 1) σ = 0.20 16.3.1 Thus, the weights of the packages of Brand B potato chips are more widely dipersed about the mean wight than re those of brand A. The mean of each random variable is given by
Binomial Distribution An important number of experiments have ( or may be viewed as having ) two outcomes. Thus, for instance in the case of a coin tossed we have two possibilities, heads or tails; When somebody is playing a game then he or she can winn or loss; If a TV is selected from a lot and tested, it could defective or not. In general experiments wth two possible outcomes are called Bernoulli or Binomial trials. It is standard practice to label one of the outcomes of the binomial experiment as succes and the other as failure. Let us give a formal definition of a binomial trial. Binomial Experiment A binomial experiment has the following properties 1 The number of trials in the experiment is fixed. 2. There are two outcomes of each trial. 3. The probability of the success in each trial is the same 4. The trial are independent of each other. The probability of success idenoted by p and the probability of failure is debnoted by q = 1 p. Example A fair die is rolled four times. Compute the probability of obtaining exaclty one 6 in the four throws. Each trial has two possible outcomes, namely: S = success = If the number falling uppermost is 6 F = failure = If the number falling uppermost is not a 6 The associated probabilities are p = 1/ 6 and q = 5 / 6 and from the multiplication principle we know that we have 2 4 = 16 possible outcomes: FFFF SFFF SSFF SSSF SSSS FSFF SFSF SSFS FFSF SFFS SFSS FFFS FSSF FSSS FSFS FFSS
Probabilty distributions and Statistics Binomial Distribution The event of obtaining exactly one succes is E = { SFFF, FSFF, FFSF, FFFS } with probability gven by Binomial Distribution In this way, if we want to find the probability of obtaining x ( = 0, 1, 2, 3, 4 ) successes in n ( = 4 ) trials is given by the expression: C(n, x ) p x q n - x P(E) = P(SFFF) + P(FSFF) + P( FFSF) + P(FFFS) = P(S)P(F)P(F)P(F) + P(F)P(S)P(F)P(F) + P( F)P(F)P(S)P(F) + P(F)P(F)P(F)P(S) = pqqq + qpqq + qqpq + qqqp Computation of Probabilities in Bernoulli ( Binomial ) Trials In a binomial experiment in which the probability of success in any trial is p, the probability of exactly x successes in n independent trials is given by = 4 p q 3 = C(4, 1 ) p q 3.386 P b ( n, x) = C(n, x ) p x q n - x Proceeding in the same way we can evaluate the probabilities of obtaining zero 6, two 6s, three 6s, and four 6s with the following results: C(4, 0 ) p 0 q 4 C(4, 2 ) p 2 q 2 C(4, 3 ) p 3 q 1 C(4, 4 )p 4 q 0 OBS We can use a function in our calculator to evaluate the above probability. To this end, press 2ND + VARS and scroll down until you obtain binompdf( then click it and you receive the function binompdf( n, p,x )
Binomial Distribution Examples 1)A fair die is rolled five times. If a 1 or a 6 lands uppermost in trial;, then the throw is considered a success. Otherwise, the throw is consider failure P (5, 3) = binompdf( 5, 1/3, 3 ).165 b P (5, 4) = binompdf( 5, 1/3, 4 ).041 b P (5, 5) = binompdf( 5, 1/3, 5 ).004 b a. Find the probabilities of obtaining exactly 0, 1, 2, 3, 4, and 5 successes, respectively, in this experiment. b. Using the results obtained in the solution to part (a), construct the binomial distribution for this experiment. The probability of success in this case is p = 2/ 6 = 1/ 3 and the probability of failure is q = 2/3. Since there are 5 trials then n = 5 and therefore 2) A fair die is rolled five times. If a 1 or a 6 lands uppermost in trial;, then the throw is considered a success. Otherwise, the throw is consider failure a. What is the probability of obtaining 0 or 1 success in the experiment? b. What is the probability of obtaining at least 1 success in the experiment? P (5, 0) = binompdf( 5, 1/3, 0 ).132 b P (5, 1) = binompdf( 5, 1/3, 1 ).329 b P (5, 2) = binompdf( 5, 1/3, 2 ).329 b a. P (5, 0) + P (5, 1) = binompdf( 5, 1/3, 0 ) b b + binompdf( 5, 1/3, 1 ) =.132 +.329 =.461.
Binomial Distribution b. P (5, 1) + P (5, 2) + P (5, 3) + P (5, 4) P (5, 5) b b b b b = binompdf( 5, 1/3, 1 ) + binompdf( 5, 1/3, 2 ) + binompdf( 5, 1/3, 3 ) + binompdf( 5, 1/3, 4 ) + binompdf( 5, 1/3, 5 ) =.868 = 1 P (5, 0) b Mean, Variance and Standard Deviation If X is a binomial random variable associated with a binomial experiment consisting of n trials with probability of success p and probability of failure q, then the mean, variance, and standard deviation are given by μ = E ( X ) = np Var ( X ) = npq σ = npq 3) A fair die is rolled five times. If a 1 or a 6 lands uppermost in a trial; then the throw is considered a success. Otherwise, the throw is consider failure. Find the mean, variance and standard deviation. μ = E ( X ) = (5 ) (1/3 ) 1.67 Var ( X ) = npq = (5 ) (1/3 ) (2/3 ) 1.11 σ = npq = (5 ) (1/3 ) (2/3 ) 1.05 If we evaluate the mean, variance, and standard deviation directly from the probaility distribution we obtain the same result. μ = E ( X ) (0 ) (. 132)+ (1 ) (. 329 )+...+(5 ) (.004 ) 1.67 Var ( X ) (0 1. 67) 2 (. 132 )+ (1 1.67 ) 2 (.329 )+...+ (5 1.67 ) 2 (.004 ) 1.1 σ 1.11 1.05
Binomial Distribution 4) A division of aron manufactures photovoltaic cells to use in the company's solar energy converters. It is estimated that 5% of the cells manufactured are defective. If a random sample of 20 is selected from a large lot of cells manufactured by the company, what is the probability that it will contain at most 2 defective cells? This is a binomial experiment with probability of success (being defective) p =.05 an probability of failureq =.95 in n = 20 trials we want to find P (20, 0) + P ( 20, 1 ) + P ( 20, 2) = b b b binompdf( 20,.05, 0 ) + binompdf( 20,.05, 1 ) + binompdf( 20,.05, 2 ). 3585 +.3774 +.1887. 9246 5) The probability that a heart transplant is succesful ( that is, the patient survives 1 year or more after undergoing such an operation ) is.7. Of six patients who have recently undergone such an operation, what is the probability that, 1 year from now, a. None of the heart recipients will be alive? b. Exactly thre will be alive? c. At least three will be alive? D All will be alive? This is a binomial experiment with n =6, p =.7, q =.3. a. P (6, 0) = binompdf( 6,.7, 0 ).. 0007. b b. P (6, 3) = binompdf( 6,.7, 3 )..1852. b c. P (6, 3) +.P (6, 4) +.P (6, 5) +.P (6, 6) = b b b b binompdf( 6,.7, 3 ) + binompdf( 6,.7, 4 ) + binompdf( 6,.7, 5 )+ binompdf( 6,.7, 6 ).9294 d. P (6, 6) = binompdf( 6,.7, 6 ).. 1176. b
Binomial Distribution Normal Distribution 6) PAR Bearings manufactures ball bearings packaged in lots of 100 each. The company's quality-control department has determined that 25 of the ball bearings manufactured do not meet the specifications imposed by a buyer. Find the average number of ball bearings per package that fail to meet the buyers' specifications. The distribuition studied previously is the type known as a finite probability distribution. Now, we will introduce the idea of a continuous probability distribution. Unlike a finite probability distribution, where we have a finite number of values or the probability, a conitunuos probability distribution is described by a function f whose domain is the set of all possible values of the random variable associated with the experiment. Such a function f is called the probability density function, and ita has the following properties: The experiment under consieration is a binomial one with probability of success ( being defective ) p =.02, therefore the expected value ( average ) is given by E(X) = np = (100)(.02) = 2 1. f(x) is nonnegative for all values of x in its domain 2. The area under the curve and above the x-axis is equal to 1.; That is, 2 ball bearing in a lot of 100 will be defective.
Normal Distribution The most important continuous probability distribution that we will study is the so called Normal Distribution. In this case the probability density function, which is bell shaped, is called a normal curve. The normal curve is completely determined by its mean μ and standard deviation σ. In fact the normal curve has the following characteristics: The probability density function has a formal expression given by : f ( x ) = and its graph is given by 1 σ 2π e 1 ( x μ ) 2 2 σ 1. The curve has a pike at x = μ 2. The curve is symmetric to the vertical line x = μ. 3. The curve always lies above the x-axis but approches to the x-axis as x extends indefinitely in either direction 4. The area under the curve is 1 5. For any normal curve, 68.27% of the area under the curve lies within 1 standard deviation from the mean value, 95.45% of the area lies within 2 standard deviations from the mean value, and 99.73% of the area lies within three standard deviations of the mean. OBS To evaluate probabilities in this case we will need to find the area under the curve of certain region, determined by the problem and to do that we will use the function normalcdf( x1, x2, μ, σ ) on our calculator which can be obtained clicking 2ND + VARS and scrolldown until we find it : normalcdf (.
Standard Normal Distribution This is the case where the mean is 0 and the standard deviation is 1. In this case the random variable use is Z. Examples 1) Let Z be the standard normal variable. Make a sketch of the appropriate region under the normal curve, and then find the values. a. P( Z < 1.24 ) b. P(0.24 < Z < 1.48 ) c. P( Z > 0.5 ) d. P ( -1.65 < Z < 2.02 ) b. P(0.24 < Z < 1.48 ) = normalcdf( 0.24, 1.48, 0, 1).34 c. P( Z > 0.5 ) = normalcdf( 0.5, EE99, 0, 1).31 d. P ( -1.65 < Z < 2.02 ) = normalcdf( -1.65, 2.02, 0, 1).93 2) Let Z be the standard normal random variable. Find the value of z if it satisfies a. P( Z < z ) =.9474 b. P( Z > z ) =.9115 c. P( -z < Z < z ) =.7888 a. P( Z < 1.24 ) = P( - < Z < 1.24 ) = normalcdf( -E99, 1.24, 0, 1).89 In this case we will use the function invnorm( A, μ, σ) giving the value of z such that the area to the left of z is equal to A.
Standard Normal Distribution a. P( Z < z ) =.9474 => z = invnorm(.9474, 0, 1) 1.62 b. P( Z > z ) =.9115 => z = invnorm( 1 -.9115, 0, 1) - 1.35 c. P( -z < Z < z ) =.7888 => -z = invnorm( (1 -.7888)/2, 0, 1) - 1.25 => z = 1.25 a. P( X < 120) = normalcdf( -E99, 120, 100, 20).8413 b. P( X > 70 ) = normalcdf( 70, E99, 100, 20).9332 c. P( 75 < X < 110 ) = normalcdf( 75, 110, 100, 20).5859 3) Suppose X is a normal random variable with µ = 100 and σ = 20. Find the values of a. P( X < 120) b. P( X > 70 ) c. P( 75 < X < 110 )
Probabilty distributions and Statistics Standard Normal Distribution Applications of the Normal Distribution This time we will look at some applications of the normal distribution. Examples 1) The medical records of infants delivered at the Kaiser Memorial Hospital show that the infants' birth weights in pounds are normally distributed with a mean of 7.4 and a standard deviation of 1.2. Find the probability than an infant selected at random from among those delivered at the hospital weighed more than 9.2 pounds at birth. Since we have a normal random variable, the probability we are looking for is P(X > 9.2) = normalcdf( 9.2, EE99, 7.4, 1.2).0668 2) Idaho Natural Produce Corporation ships potatoes to its distributors in bags whose weights are normally distributed with a mean weight of 50 pounds and standard deviation of 0.5 pound. If a bag of potatoes is selected at random from a shipment, what is the probability that it weighs: a. More than 51 pounds? b. Less than 49 pounds? c. Between 49 and 51 pounds? In this case the probabilities are given by a. P(X > 51) = normalcdf( 51, EE99, 50, 0.5).0228 b. P(X < 49) = normalcdf( -EE99, 49, 50, 0.5).0228 c. P( 49 < X < 51) = normalcdf( 49, 51, 50, 0.5).9544
Standard Normal Distribution Applications of the Normal Distribution 3) The grade point average (GPA) of the senior class of Jefferson High School is normally distributed with a mean of 2.7 and standard deviation of 0.4. If a senior in the top 10% of his or her is elegible for admission t any of the nine campuses of the state university system, what is the minimum GPA that a senior should have to ensure elegibility for university admission? Approximating Binomial Distributions Suppose we are given a binomial distribution associated with a binomial experiment involving n trials, each with a probability of success p and probability of q. Then, if n is large and p is not close to 0 or 1, the binomial distribution may be approximated by a normal distribution with μ = np σ = npq In this case we are looking for the value x of the random variable X, normally distributed, such that P( X < x ) =.9 = > x = invnorm(.9, 2.7, 0.4) 3.2 Example 1) An automobile manufacturer receives the microprocessors that are used to regulate fuel consumption in its automobile in shipments of 1000 each from a certain supplier. It has been estimated that, on the average, 1% of the microprocessors manufactured by the supplier are defetive. Determine the probability that more than 20 of the microprocessors in a single shipment are defective?
Standard Normal Distribution Applications of the Normal Distribution Let X denote the number of defective microprocessors in a single shipment. Then X has abinomial distribution with n= 1000, p =.01, and q =.99, so μ = (1000 ) (. 01 ) = 10 σ = (1000 ) (.01 ) (. 99 ). 3.15 P b ( X > 20 ) = P b ( X 21 ) P N ( Y > 20.5 ) = normalcdf( 20.5, EE99, 10, 3.15).0004 2) The probability that a heart transplant is succesful ( that is, the patient survives 1 year or more after undergoing such an operation ) is.7. Of 100 patients who have recently undergone such an operation, what is the probability that, 1 year from now. What is the probability that : a. Fewer than 75 will survive 1 year or more after the operation? b. Between 80 and 90, inclusive, will survive 1 year or more after the operation?. Let X denote the number of patients who survived 1 year or more after undergoing a heart transplant at the Medical Center. Then X has a binomial distribution with n= 1000, p =.7, and q =.3, so μ = (1000) (.7 ) = 70 σ = (1000 ) (. 7 ) (.3 ) 4.58 a. P b ( X < 75 ) = P b ( X 74 ) P N ( Y < 74.5 ) = normalcdf( -EE99, 74.5, 70, 4.58).8365 b. P b ( 80 X 90 ) P ( 79.5 < Y < 90.5 ) N = normalcdf( 79.5, 90.5, 70, 4.58).0192