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 [tex]\overline X[/tex] = 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
[tex]P(X < 31.6) = P(\dfrac{\overline X - \mu}{\dfrac{\sigma }{\sqrt{n}}}< \dfrac{\overline X - \mu}{\dfrac{\sigma }{\sqrt{n}}})[/tex]
[tex]P(X < 31.6) = P(\dfrac{31.6 - \mu}{\dfrac{\sigma }{\sqrt{n}}}< \dfrac{31.6 - 31}{\dfrac{1.9 }{\sqrt{39}}})[/tex]
[tex]P(X < 31.6) = P(Z< \dfrac{31.6 - 31}{\dfrac{1.9 }{\sqrt{39}}})[/tex]
[tex]P(X < 31.6) = P(Z< \dfrac{0.6}{\dfrac{1.9 }{6.245}})[/tex]
[tex]P(X < 31.6) = P(Z< 1.972)[/tex]
From standard normal tables
P(X < 31.6) = 0.9756
For sample size n = 76
[tex]P(X < 31.6) = P(\dfrac{\overline X - \mu}{\dfrac{\sigma }{\sqrt{n}}}< \dfrac{\overline X - \mu}{\dfrac{\sigma }{\sqrt{n}}})[/tex]
[tex]P(X < 31.6) = P(\dfrac{31.6 - \mu}{\dfrac{\sigma }{\sqrt{n}}}< \dfrac{31.6 - 31}{\dfrac{1.9 }{\sqrt{76}}})[/tex]
[tex]P(X < 31.6) = P(Z< \dfrac{31.6 - 31}{\dfrac{1.9 }{\sqrt{76}}})[/tex]
[tex]P(X < 31.6) = P(Z< \dfrac{0.6}{\dfrac{1.9 }{8.718}})[/tex]
[tex]P(X < 31.6) = P(Z< 2.75)[/tex]
From standard normal tables
P(X < 31.6) = 0.9970
