Question

Suppose that the mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is 65 mm and that the population standard deviation is 5 mm.

(a) If the distribution of interpupillary distance is normal and a random sample of n = 25 adult males is to be selected, what is the probability that the sample mean distance x for these 25 will be between 63 and 67 mm? (Round all your intermediate calculations to four decimal places. Round the answers to four decimal places.)
P =

What is the probability that the sample mean distance x for these 25 will be at least 68 mm?
P =

(b) Suppose that a sample of 100 adult males is to be obtained. Without assuming that interpupillary distance is normally distributed, what is the approximate probability that the sample mean distance will be between 63 and 67 mm? (Round all your intermediate calculations to four decimal places. Round the answers to four decimal places.)
P =

Without assuming that interpupillary distance is normally distributed, what is the approximate probability that the sample mean distance will be at least 68 mm?
P =

Answer 1

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	65
std deviation   =σ=	5.000
sample size       =n=	25
std error=σx̅=σ/√n=	1.000

  probability that the sample mean distance x for these 25 will be between 63 and 67 mm:

probability =P(63<X<67)=P((63-65)/1)<Z<(67-65)/1)=P(-2<Z<2)=0.9772-0.0228=0.9544

probability that the sample mean distance x for these 25 will be at least 68 mm:

probability =P(X>68)=P(Z>(68-65)/1)=P(Z>3)=1-P(Z<3)=1-0.9987=0.0013

b)

sample size       =n=	100
std error=σx̅=σ/√n=	0.50000

probability that the sample mean distance will be between 63 and 67 mm:

probability =P(63<X<67)=P((63-65)/0.5)<Z<(67-65)/0.5)=P(-4<Z<4)=1-0=1.0000

approximate probability that the sample mean distance will be at least 68 mm:

probability =P(X>68)=P(Z>(68-65)/0.5)=P(Z>6)=1-P(Z<6)=1-1=0.0000