Question

3. MNM Corporation gives each of its employees an aptitude test. The scores on the test are normally distributed with a mean of 75 and a standard deviation of 15. A simple random sample of 25 is taken from a population of 500. What are the expected value, the standard deviation, and the shape of the sampling distribution of x? What is the probability that the average aptitude test in the sample will be between 70.14 and 82.14? a. b. c. What is the probability that the average aptitude test in the sample will be greater than 82.68? d. What is the probability that the average aptitude test in the sample will be less than 78.69? e. Find a value, C, such that P/x 2C)-015

Business and Economic Statistics

Answer 1

Population has mean and standard deviation as μ 75 and σ 15 respectively. The sample size is n -25 a) Calculate the mean and standard deviation of the sampling distribution as follows: 11,-75, sample mean is as unbiased estimator of population mean. The sample variance is as follows: Sample_ Population 7l 25-(15s) 25 216 Sample Sample /216 14.697 b) Calculate the probability that the average aptitude test in the sample will be between 70.14 and 82.14 as follows P(70.14 s s 82.14) P(r s 82.14)-P( 70.14) -pl x-μ < 82.14-75 pl x_μ 70.14-75 an 14.69//25n 14.697/25 ) n 14.697 / 25 - P(Zs 2.4291)- P(Z<-1.6534 0.9924-0.0491 - 0.943:3 Therefore, the probability that the average aptitude test in the sample will be between 70.14 and 82.14 is 0.9433

c) Calculate the probability that the average aptitude test in the sample will be greater than 82.68 as follows: X- σ /、In 82.68-75 14.697/ 25 P(x>82.68) 1-P = 1-P(z 2.6128) 1-0.9955 0.0045 Therefore, the probability that the average aptitude test in the sample will be greater than 82.68 is 0.0045. d) Calculate the probability that the average aptitude test in the sample will be less than 78.69 as follows: 78.69-75 14.697/'/25 x- P(x78.69) P n - P(Z 1.2554) 0.8953 Therefore, the probability that the average aptitude test in the sample will be less than 78.69 is 0.8953. e) Find a value, C as follows: r-75 0.015 n 14.697/ 25 0.015 2.9394 Observe that the value corresponding to 0.015 cumulative probabilities is -2.1701.

x-75 2.9394 2.1701 x-75--6.3788 ' 68.6212 X s