Question

A random sample is drawn from a normally distributed population with mean μ = 31 and standard deviation σ = 1.9. Calculate the probabilities that the sample mean is less than 31.6 for both sample sizes

Answer 1

Answer:

For sample size n = 39 ; P(X < 31.6) = 0.9756

For sample size n = 76 ; P(X < 31.6) = 0.9970

Step-by-step explanation:

Given that:

population mean μ = 31

standard deviation σ = 1.9

sample mean  $\overline X$ = 31.6

Sample size n                 Probability

39

76

The probabilities that the sample mean is less than 31.6 for both sample size can be computed as  follows:

For sample size n = 39

$P(X < 31.6) = P(\dfrac{\overline X - \mu}{\dfrac{\sigma }{\sqrt{n}}}< \dfrac{\overline X - \mu}{\dfrac{\sigma }{\sqrt{n}}})$

$P(X < 31.6) = P(\dfrac{31.6 - \mu}{\dfrac{\sigma }{\sqrt{n}}}< \dfrac{31.6 - 31}{\dfrac{1.9 }{\sqrt{39}}})$

$P(X < 31.6) = P(Z< \dfrac{31.6 - 31}{\dfrac{1.9 }{\sqrt{39}}})$

$P(X < 31.6) = P(Z< \dfrac{0.6}{\dfrac{1.9 }{6.245}})$

$P(X < 31.6) = P(Z< 1.972)$

From standard normal  tables

P(X < 31.6) = 0.9756

For sample size n = 76

$P(X < 31.6) = P(\dfrac{\overline X - \mu}{\dfrac{\sigma }{\sqrt{n}}}< \dfrac{\overline X - \mu}{\dfrac{\sigma }{\sqrt{n}}})$

$P(X < 31.6) = P(\dfrac{31.6 - \mu}{\dfrac{\sigma }{\sqrt{n}}}< \dfrac{31.6 - 31}{\dfrac{1.9 }{\sqrt{76}}})$

$P(X < 31.6) = P(Z< \dfrac{31.6 - 31}{\dfrac{1.9 }{\sqrt{76}}})$

$P(X < 31.6) = P(Z< \dfrac{0.6}{\dfrac{1.9 }{8.718}})$

$P(X < 31.6) = P(Z< 2.75)$

From standard normal  tables

P(X < 31.6) = 0.9970