Answer to Question #194647 in Statistics and Probability for Liche Kelvin

Question

(b) Time spent studying by students in the week before the sessional examinations follows a normal distribution with a population standard deviation of eight hours. A random sample of 16 students is taken in order to estimate the mean study time for the population of all students. What is the probability that the difference between the sample mean and the population mean

(i) exceeds two hours?

(ii) is less than three hours?

(c) From past experience, the Software Manager at the Northrise University Computer Centre has found that the downtime (in hours) per day of the main server follows a normal distribution with a population variance of 0.25. If a random sample of 21 days is selected and the data on the downtime are recorded, what is the probability that the sample variance is between 0.092875 and 0.469625?

Answer 1

(b) $\sigma=8, n=16$

Let $\mu$ and x be the population and sample mean respectively-

(i) Probability that the difference between the sample mean and the population mean exceeds two hours-

$P(x-\mu>2)=P(z>\dfrac{2}{\frac{8}{\sqrt{16}}})=P(z>\dfrac{2\times 4}{8})=P(z>1)=$ 0.15866

(ii) Probability that the difference between the sample mean and the population mean is less than three hours-

$P(x-\mu<3)=P(z<\dfrac{3}{\frac{8}{\sqrt{16}}})=P(z<\dfrac{3\times 4}{8})=P(z<1.5)=$ 0.93319

(c) Population variance $\sigma^2=0.25$

Population standard deviation $\sigma=\sqrt{0.25}=0.5$

n=21,

Probability that Sample variance is between 0.092875 and 0.469625 is-

$P(0.092875<X<0.469625)=P(\dfrac{0.092875-0.25}{\frac{0.5}{\sqrt{21}}}<z<\dfrac{0.469625-0.25}{\frac{0.5}{\sqrt{21}}})$

$=P(-1.44<z<2.012)=0.90296$