Question

5. 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. a. What are the expected value, the standard deviation, and the shape of the sampling distribution of x? b. What is the probab]lity that the average aptitude test in the sample will be between 70.14 and 82.14? 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? Find a value, C, such that P(x > C)=.015. e.

Answer 1

Solution . Here , Expected valueft = 75 any. standand deviation, rais (Any) The which sampling distribution of t has bell shape curve is symmetrucal also you can say it has symmetrical bell shepe cuave. Like :- on (6) The standardized normal random Variable is given by 2= x- Let x be random Variable for the scores of the aptitude test. P (70.14<XL 82.14) =P [10:13-154 70.14-15, X-WL 82.14-75 15 Note: $ €2) = 1-0 () p(-0.324 < Z < 0.476) $ (0.476)-0(-0.324) $10.476)-(-460.324)] $16.476)+ $(0.324) - 1 You can from donmal Table. " get prz) values 0.6808 + 0.6255 -1 L(Any) 0.3063 (C) P(x > 82.68 ) = 1-PCX5 82.68) - L- P(Z 4 *R.53-75) 1- P(Z < 0.512) P(0.512) > I-0.6950 I (Ang). - 10.305
q
id) PCX < 78.69) - P(Z < 78.69-75 15 P(Z < 0.0 -0.246) 0.5981 I (Ang). le) P(x2e): 0.015 then Co ?? I- - PCX?C)= 0.015 L-0.015 P(246) P() Normal table you will 0.985 = c-ť 2.19 I get a for given $(2) → C- (2.190th Q.199x15 + 75 - 322.85 +75 => 107.85 (Ang) Thanks you