1. Find the area under the standard normal distribution curve for each. (a) Between z = 0.19 and z = 1.23.

Transcription

1 Statistics Test 3 Name: KEY 1 Find the area under the standard normal distribution curve for each a Between = 019 and = 123 normalcdf 019, 123 = b To the left of = 156 normalcdf 100, 156 = c To the right of = 038 normalcdf 038, 100 = Find each probability using the standard normal distribution curve for each a P 009 < < 242 = normalcdf 009, 242 = b P > 168 = normalcdf 168, 100 = c P < 023 = normalcdf 100, 023 = 05910

2 3 Find the indicated score The graph depicts the standard normal distribution with mean 0 and standard deviation 1 a = invnorm08980 = 127 b = invnorm = Find the critical value of a 012 = invorm1 012 = 117 b 008 = invorm1 008 = 141

3 5 Find the critical value of that represents the 45th percentile = invnorm045 = The waiting time in line at a Starschmuchs Coffee is normally distributed with a mean of 32 minutes and a standard deviation of 13 minutes Find the probability that a randomly selected customer has to wait a Less than 1 minute µ = 32 and = 13 Let = the continuous random variable CRV representing a randomly selected wait time We convert the distribution of values over to its distribution of scores, and find the corresponding area under the standardied normal curve Normal Distribution of Wait Times P < 1 = P < µ = P < = P < 169 = normalcdf 100, 169 = Alternatively, we could just find the area under the distribution of wait times,, with P < 1 = normalcdf,, µ, = normalcdf 10 10, 1, 32, 13 = b more than 2 minutes P > 2 = P > µ = P > = P > 092 = normalcdf 092, 100 = Normal Distribution of Wait Times Alternatively, we could just find the area under the distribution of wait times,, with P > 2 = normalcdf,, µ, = normalcdf2, 10 10, 32, 13 = 08220

4 c between 075 minutes and 2 minutes P 075 < < = P 13 = P 188 < < 092 < < Normal Distribution of Wait Times = normalcdf 188, 092 = P 075 < < 2 = normalcdf075, 2, 32, 13 = The average yearly precipitation in San Diego is 962 inches with a standard deviation of 442 inches and precipitation amounts are normally distributed a b Find the probability that a randomly selected year will have precipitation greater than 12 inches µ = 962 in and = 442 in Let = the continuous random variable CRV representing a randomly selected yearly precipitation amount P > 12 = P > µ = P > = P > = normalcdf054, 100 = Normal Distribution of Precipitations Alternatively, we could just find the area under the distribution of precipitations,, with P > 12 = normalcdf,, µ, = normalcdf12, 10 10, 962, 442 = Find the probability that five randomly selected years will have an average precipitation greater than 8 inches Let = the continuous random variable CRV representing a randomly selected sample mean yearly precipitation amount Sampling Distn of Sample Means P > 8 = P > µ / n = P > / 5 µ = 962 and = 442/ 5 = 198

5 = P > 082 = normalcdf 082, 100 = Alternatively, we could just find the area under the sampling distribution of sample mean precipitations,, with P > 8 = normalcdf,, µ, = normalcdf8, 10 10, 962, 442/ 5 = c cont Find the precipitation amount from the distribution of precipitations that represents the 75th percentile precipitation, = invnormpercentile, µ, = invnorm075, 962, 442 = 126 in Alternatively, we first find the value from the standard normal distribution that corresponds to the 75th percentile = invnorm075 = Second, we solve = µ = µ + for to get Afterwards, replace µ with 962, with and with 442 so that = = 126 in 8 Some passengers died when a water taxi sank in Baltimore s inner harbor Men are typically heavier than women and children, so when loading a water taxi, let s assume a worst-case scenario in which all passengers are men Based on data from the National Health and Nutrition Survey, assume that weights of men are normally distributed with a mean of 172 lb and a standard deviation of 29 lb a Find the probability that if an individual man is randomly selected, his weight will be greater than 175 lb µ = 172 lb and = 29 lb Let = the continuous random variable CRV representing the weight of a randomly selected man

6 Distn of Men s Weights, P > 175 = P > µ = P > = P > = normalcdf010, 100 = b Alternatively, we could just find the area under the distribution of weights,, with P > 175 = normalcdf,, µ, = normalcdf175, 10 10, 172, 29 = Find the probability that 20 men will have a mean weight that is greater than 175 lb so that their total weight exceeds the safe capacity of 3500 lb Let = the continuous random variable CRV representing a randomly selected sample mean weight P > 175 = P > µ / n = P > 29/ 20 = P > 046 Sampling Distn of Sample Means µ = 172 and = 29/ 20 = 65 = normalcdf046, 100 = Alternatively, we could just find the area under the sampling distribution of sample mean weights,, with P > 175 = normalcdf,, µ, = normalcdf175, 10 10, 172, 29/ 20 = The average per capita spending on health care in the United States is $5274 The standard deviation is $600 and the distribution of health care spending is approximately normal Find the limits of the middle 50% of individual health care expenditures

7 Let be the continuous random variable CRV representing a randomly selected individual health care expenditure Find the scores from the standard normal distribution corresponding to the 25th and 75th percentiles Then use = µ = invnorm025 = , and by symmetry, 025 = x low = µ+ = $ $600 $ x high = µ+ = $ $600 $ Alternatively, we could bypass using the standard normal distribution of scores with: x low = invnormpercentile, µ, = invnorm025, 5274, 600 = $ x high = invnormpercentile, µ, = invnorm075, 5274, 600 = $ A prestigious college decides to only take applications from student who have scored in the top 5% on the SAT test The SAT scores are approximately normally distributed with a mean of 490 and a standard deviation of 70 Find the score that is necessary to obtain in order to qualify for applying to this college Let be the continuous random variable CRV representing SAT scores A student must score in the 95th or higher percentile in order to be admitted First, we find the value from the standard normal distribution that corresponds to the 95th percentile Second, we solve = µ = µ + for to get Afterwards, replace µ with 490, with and with 70 so that = = 605 An alternate solution route is = invnorm095 = = invnormpercentile, µ, = invnorm095, 490, 70 = 605

8 11 Americans ate an average of 257 pounds of Krusty-O Cereal each last year and spent an average of $6150 per person doing so If the standard deviation for consumption is 375 pounds and the standard deviation for the amount spent is $589, find the following: a b The probability that the sample mean Krusty-O cereal consumption for a random sample of 40 American consumers exceeded 27 pounds Let = the continuous random variable CRV representing a randomly selected sample mean consumption amount P > 27 = P > µ / n = P > = P > / 40 = normalcdf219, 100 = Sampling Distn of Sample Means µ = 257 lb and = 375/ 40 = 059 lb Std Normal Distn Alternatively, we could just find the area under the sampling distribution with P > 27 = normalcdf,, µ, = normalcdf27, 10 10, 257, 375/ 40 = The probability that for a random sample of 50, the the average yearly amount spent on Krusty-O Cereal was between $6000 and $100 Let = the continuous random variable CRV representing a randomly selected sample mean yearly amount spent P 60 < < 100 µ = P / n < < µ / n = P 589/ < < 589/ 50 Sampling Distn of Sample Means µ = $6150 and = 589/ 50 = $083 = P 18 < < 4622 = normalcdf 18, 4622 = Std Normal Distn Alternatively, we could just find the area under the sampling distribution with P 60 < < 100 = normalcdf60, 100, 6150, 589/ 50 = 09641

9 12 Use the normal approximation of the binomial probability distribution to find the probabilities for the discrete random variable, a Find the probability that is 19, assuming n = 40 and p = 05 We find that µ = n p = = 20, = npq = 10, and both np 5 and nq 5 Thus, EACT SOLUTION Binomial Probability Distribution P = 19 = binomp df40, 05, 19 = TABLE APPROIMATION P = 19 = P 185 < < 195 µ = P = P < < 10 < < µ = P 047 < < 016 = normalcdf 047, 016 = TI 84/83 + CALCULATOR APPROIMATION P = 19 = normalcdf185, 195, 20, 10 = 01196

10 b Find the probability that is 3, assuming n = 25 and p = 04 We find that µ = n p = = 10, = npq = 6, and both np 5 and nq 5 EACT SOLUTION P = 3 = binompdf25, 04, 3 = Binomial Probability Distribution TABLE APPROIMATION P = 3 = P 25 < < 35 µ = P P < < 6 < < µ = P 306 < < 265 = normalcdf 306, 265 = TI 84/83 + CALCULATOR APPROIMATION P = 3 = normalcdf25, 35, 10, 6 = 00029

11 c Find the probability that is at least 15, assuming n = 30 and p = 05 We find that µ = n p = = 15, = npq = 75, and both np 5 and nq 5 EACT SOLUTION Binomial Probability Distribution P 15 = 1 binomcdf30, 05, 14 = TABLE APPROIMATION P 15 = P > 145 = P > µ = P > = P > 018 = normalcdf 018, 100 = TI 84/83 + CALCULATOR APPROIMATION P 15 = normalcdf145, 10 10, 15, 75 = 05724

12 13 Use the normal approximation of the binomial probability distribution to find the probabilities for the discrete random variable, a Find the probability that is fewer than 5, assuming n = 300 and p = 007 We find that µ = n p = = 21, = npq = 1953, and both np 5 and nq 5 Binomial Probability Distribution EACT SOLUTION P < 5 = P 4 = binomcdf300, 007, 4 = E = TABLE APPROIMATION P < 5 = P < 45 = P < µ = P < = P < 373 = normalcdf 100, 373 = TI 84/83 + CALCULATOR APPROIMATION P < 5 = normalcdf 10 10, 45, 21, 1953 =

13 b Find the probability that is at most 8, assuming n = 100 and p = 013 We find that µ = n p = = 13, = npq = 1131, and both np 5 and nq 5 EACT SOLUTION Binomial Probability Distribution P 8 = binomcdf100, 013, 8 = TABLE APPROIMATION P 8 = P < 85 = P < µ = P < = P < 133 = normalcdf 100, 133 = TI 84/83 + CALCULATOR APPROIMATION P 8 = normalcdf 10 10, 85, 13, 1131 = 00904

14 c Find the probability that is more than 35, assuming n = 50 and p = 06 We find that µ = n p = = 30, = npq = 12 = 2 3, and both np 5 and nq 5 Binomial Probability Distribution EACT SOLUTION P > 35 = 1 binomcdf50, 06, 35 = TABLE APPROIMATION P > 35 = P > 355 = P > µ = P > = P > 159 = normalcdf159, 100 = TI 84/83 + CALCULATOR APPROIMATION P > 35 = normalcdf355, 10 10, 30, 6 = 00562

15 14 Use the normal approximation of the binomial probability distribution Two out of five adult smokers acquired the habit by age 14 If 400 smokers are randomly selected, find the probability that 170 or fewer acquired the habit by age 14 We determine that p = 2 = 04 and n = 400 Let be the discrete random variable 5 representing the number of smokers out of 400 who acquired the habit by age 14 Then, µ = n p = = 160, = npq = 96 = 4 6, and both np 5 and nq 5 Binomial Probability Distribution EACT SOLUTION P 170 = binomcdf400, 04, 170 = TABLE APPROIMATION P 170 = P < 1705 = P < µ = P < = P < 107 = normalcdf 100, 107 = TI 84/83 + CALCULATOR APPROIMATION P 170 = normalcdf 10 10, 1705, 160, 96 = 00029

16 15 Use the normal approximation of the binomial probability distribution According to Mars the candy company, 24% of M&Ms plain candies are blue Assuming that the claimed blue M&Ms rate of 24% is correct, find the probability of randomly selecting 100 M&Ms and getting at most 20 that are blue We determine that p = 024 and n = 100 Let be the discrete random variable representing the number of blue M&Ms out of 100 Then, µ = n p = = 24, = npq = 1824, and both np 5 and nq 5 Binomial Probability Distribution EACT SOLUTION P 20 = binomcdf100, 024, 20 = TABLE APPROIMATION P 20 = P < 205 = P < µ = P < 1824 = P < 082 = normalcdf 100, 082 = TI 84/83 + CALCULATOR APPROIMATION P 20 = normalcdf 10 10, 205, 24, 1824 = 02062

17 16 Find the critical value α/2 that corresponds to a 92% confidence interval A 92% Confidence interval means that 092 = 1 α, so that α = 008 Then, and α/2 = 008/2 = 004 α/2 = 004 = invnorm1 α/2 = invnorm096 = First-semester GPAs for a random selection of freshmen at a large university are shown below Estimate the true mean GPA of the freshman class with 99% confidence Assume = 062 and that the distribution of first-semester GPAs is normal We find the sample mean is = and 99% confidence implies α/2 = 0005 = invnorm = 258 Then, α/2 n or < µ < + α/2, n < µ < , or 258 < µ < 324

18 18 Find the critical value t α/2 that corresponds to a 90% interval, assuming n = 10 A 90% Confidence interval means that 090 = 1 α, so that α = 010 Then, and α/2 = 010/2 = 005 t α/2 = t 005 = invt 1 α/2, n 1 = invt 095, 9 = The approximate costs in thousands for a 30-second spot for various cable networks in a random selection of cities are shown below Estimate the true population mean cost for a 30-second advertisement on cable network with 90% confidenceassume the population of costs is approximately normal We find the sample mean is = , sample standard deviation is s = and 90% confidence implies t α/2 = t 005 = invt 095, 16 = Then, s s t α/2 < µ < + t α/2, n n or < µ < , or 320 < µ < 709

19 20 A university dean of students wishes to estimate the average number of hours students spend doing homework per week The standard deviation from a previous study is 62 hours How large a sample must be selected if he wants to be 99% confident of finding whether the true mean differs from the sample mean by 15 hours? We are asked to determine the sie, n, of a sample necessary for an interval estimate of the average weekly study amount The formula is α/2 2 n = E We are told to assume = 62 and E = 15 If 99% Confidence is desired, then 099 = 1 α, or α = 001 This implies we should use α/2 = 0005 = invnorm0995 = 258 Then, α/ = n = = 114 E Thirty randomly selected students took the calculus final If the sample mean was 95 and the standard deviation was 66, construct a 99% confidence interval for the mean score of all students We are not given, so we use a t distribution We are given n = 30, = 95 and s = 66 99% confidence implies 099 = 1 α, or α = 001, and t α/2 = t 0005 = invt 0995, 29 = Then, s s t α/2 < µ < + t α/2, n n or < µ < , or 92 < µ < 98

20 22 A study of 35 golfers showed that their average score on a particular course was 92 The standard deviation of the population is 5 Find the 95% confidence interval of the mean score for all golfers We are given = 5, so we use a distribution We are given n = 35 and = 92 95% confidence implies 095 = 1 α, or α = 005, and α/2 = 0025 = invnorm0975 = 196 Then, α/2 n < µ < + α/2, n or < µ < , or 903 < µ < A recent study of 75 workers found that 53 people rode the bus to work each day Find the 95% confidence interval of the proportion of all workers who rode the bus to work We are given n = 75 and = 53, so the sample proportion is ˆp = % confidence implies 095 = 1 α, or α = 005, and α/2 = 0025 = invnorm0975 = 196 Then, ˆp α/2 or ˆp ˆq ˆp ˆq n < p < ˆp + α/2 n, < p < , 75 or 060 < p < 081

21 24 It is believed that 25% of US homes have a direct satellite television receiver How large a sample is necessary to estimate the true population of homes which do with 95% confidence and within 3 percentage points? How large a sample is necessary if nothing is known about the proportion? n = ˆp ˆq α/2 2 E 2 = = 801 Assuming nothing is known about the proportion, n = 052 α/2 2 E 2 = = A recent poll showed results from 2000 professionals who interview job applicants 26% of them said the biggest interview turnoff is that the applicant did not make an effort to learn about the job or the company A 95% confidence interval estimate was used and the margin of error was ±3 percentage points Describe what is meant by the statement the margin of error was ±3 percentage points When using 26% to estimate the value of the population percentage, the maximum likely difference between 26% and the true population percentage is three percentage points, so the interval from 23% to 29% is likely to contain the true population percentage The Sampling Distribution of Sample Proportions, ˆp