Question

Let X ¯ ^ be the mean of a random sample of size 50 drawn from a population with mean 112 and standard deviation 40.

Determine the expected value of the sample mean, E { X ¯ ^ } ., the standard deviation of the sample mean, the probability that the sample mean assumes a value greater than 113, and the probability that the sample mean assumes a value between 110 and 114

Answer 1

Using Central Limit theorem, we know,

$\overline{X}\sim N(\mu,\frac{\sigma^{2}}{n})\sim N(112,\frac{40^{2}}{50}) \sim N(112,32)$

a)

Expected value = $E(\overline{X})=112$

b)

Standard deviation = $\sqrt{V(\overline{X})}=\sqrt{32}=5.6569$

c)

Required probability =

$\\P(\overline{X}>113)=P(Z>\frac{113-112}{\sqrt{32}})=P(Z>0.18)=1-P(Z<0.18)\\ \\ =1-0.57142=0.42858$

d)

Required probability =

$\\P(110<\overline{X}<114)=P(\frac{110-112}{\sqrt{32}}<Z<\frac{114-112}{\sqrt{32}})\\ \\ =P(-0.35<Z<0.35)=P(Z<0.35)-P(Z<-0.35)\\ \\ =P(Z<0.35)-P(Z>0.35)=P(Z<0.35)-[1-P(Z<0.35)]\\ \\ =0.63683-[1-0.63683]=0.27366$