Question

The heights of 1000 students are approximately
normally distributed with a mean of 174.5 centimeters
and standard deviation is unknown. Suppose
200 random samples of size 25 are drawn from this population
and the means recorded to the nearest tenth of
a centimeter.

with n=6 and assume the true σ is unknown and S=7.5 centimeters instead

Determine
(a) the mean and standard deviation of the sampling
distribution of ¯X ;
(b) the number of sample means that fall between 172.5
and 175.8 centimeters inclusive;
(c) the number of sample means falling below 172.0
centimeters.

Answer 1

(a) the mean and standard deviation of the sampling
distribution of ¯X ;

Mean of the sampling distribution of Xbar = µ = 174.5

Estimate for the standard deviation of the sampling distribution of Xbar = S/sqrt(n) = 7.5/sqrt(6) = 3.06186

(b) the number of sample means that fall between 172.5
and 175.8 centimeters inclusive;

Here, we have to find P(172.5<Xbar<175.8)

P(172.5<Xbar<175.8) = P(Xbar<175.8) – P(Xbar<172.5)

Find P(Xbar<175.8)

Z = (Xbar - µ_xbar)/σ_xbar

Z = (175.8 - 174.5) / 3.06186

Z = 0.42458

P(Z<0.42458) = P(Xbar<175.8) = 0.664428

(by using z-table)

Now, find P(Xbar<172.5)

Z = (172.5 - 174.5) / 3.06186

Z = -0.6532

P(Z<-0.6532) = P(Xbar<172.5) = 0.256814

(by using z-table)

P(172.5<Xbar<175.8) = P(Xbar<175.8) – P(Xbar<172.5)

P(172.5<Xbar<175.8) = 0.664428 - 0.256814

P(172.5<Xbar<175.8) = 0.40761

Required probability = 0.40761

(c) the number of sample means falling below 172.0
centimeters.

First we have to find P(Xbar<172)

Z = (Xbar - µ_xbar)/σ_xbar

Z = (172 - 174.5) / 3.06186

Z = -0.8165

P(Z<-0.8165) = P(Xbar<172) = 0.207108

(by using z-table)

Number of sample means falling below 172.0 = N*p = 200*0.207108 = 41.4216

Number of sample means falling below 172.0 = 41 approximately