Question

A fair die is rolled 500 times. Find the probability that a number 2 or less comes up on at most 150 rolls. b. The scores on a psychological profile test given to job applicants at a nuclear facility are known to be normally distributed with a mean of 65 and a standard deviation of 10. 1. What score is required for an applicant to be in the top 10%? 2. Suppose a random sample of 16 applicants is selected, find the probability that the sample mean score is above 70.

Answer 1

1)

n=	500	p=	1/6
here mean of distribution=μ=np=		83.3333
and standard deviation σ=sqrt(np(1-p))=		8.3333

for normal distribution z score =(X-μ)/σx
therefore from normal approximation of binomial distribution and continuity correction:

probability that a number 2 or less comes up on at most 150 rolls=P(X<=150)=P(Z<(150-83.33)/8.33)

=P(Z<8.06)=0.0000

b)

1)

for 90th percentile critical value of z=		1.28
therefore corresponding value=mean+z*std deviation=			77.80

2)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	65
std deviation   =σ=	10.0000
sample size       =n=	16
std error=σx̅=σ/√n=	2.5000

probability =

P(X>70)

=

P(Z>2)=

1-P(Z<2)=

1-0.9772=

0.0228