Answer to Question #303832 in Statistics and Probability for James

Question

As studied, the average number of hours spent by senior high school students for their online classes a week is 25 hours with a standard deviation of 4 hours. Assuming that the study is is true and the data is normally distributed.

Answer 1

Let $X=$ the number of hours spent by senior high school students for their online classes a week: $X\sim N(\mu, \sigma^2)$

Then $Z=\dfrac{X-\mu}{\sigma}\sim N(0, 1)$

Given $\mu=25\ h, \sigma=4\ h.$

a.

$$P(26<X<29)=P(X<29)-P(X\le26)$$

$$=P(Z<\dfrac{29-25}{4})-P(Z\leq\dfrac{26-25}{4})$$

$$=P(Z<1)-P(Z\leq0.25)$$

$$\approx0.8413447-0.5987063\approx0.2426$$

b.

$$P((X<21)\cup(X>30))$$

$$=P(X<21)+1-P(X\leq30)$$

$$=1+P(Z<\dfrac{21-25}{4})-P(Z\leq\dfrac{30-25}{4})$$

$$=1+P(Z<-1)-P(Z\leq1.25)$$

$$\approx1+0.158655-0.894350\approx0.2643$$

с.

$$\bar{X}\sim N(\mu, \sigma^2/n)$$

Given $n=12$

$$P(\bar{X}>24)=1-P(\bar{X}\le 24)$$

$$=1-P(Z\le\dfrac{24-25}{4/\sqrt{12}})$$

$$\approx1-P(Z\le-0.866025)$$

$$\approx0.8068$$