Question

Please computer typed, do not use hand writing! ! ! Thanks

Question 2 (4 marks) Part a) A sample of n-25 observations is drawn from a normal population with μ-100 and o-20. Find the following. i) P(X<96) ii) P(96-X-105) Part b) The amount of time the university professors devote to their jobs per week is normally distributed with a mean of 52 hours and a standard deviation of 6 hours. i) What is the probability that a professor works for more than 60 hours per weeks? ii) Find the probability that the mean amount of work per week for three randomly selected professors is more than 60 hours?

Answer 1

Part(a)

Here n= 25,  $\mu$ = 100 , $\sigma$= 20

$Z= \frac{\bar{x}-\mu }{\sigma /\sqrt{n}} = \frac{96-100 }{20 /\sqrt{25}} = -1$

By using standard normal tables

1) here P($\bar{x}$< 96) = P(Z < -1) = 0.1587

2) for 105

$Z= \frac{\bar{x}-\mu }{\sigma /\sqrt{n}} = \frac{105-100 }{20 /\sqrt{25}} = 1.25$

Now P(96 < $\bar{x}$ <105) = P(-1 < Z< 1.25)

= P(Z<1.25) -P(Z<-1)

= 0.8944 - 0.1587

= 0.7357

Part(b)

Here n= 25, $\mu$ =100, $\sigma$= 20

$Z= \frac{\bar{x}-\mu }{\sigma } = \frac{60-52}{6 } = 1.333$

1) P(works for more than 60 hours) = P( x>60)

= P(Z > 1.333)

= 1- P(Z < 1.333)

= 1-0.9087

= 0.0913

2) for 3 proffessors n=3 , then

$Z= \frac{\bar{x}-\mu }{\sigma/\sqrt{n} } = \frac{60-52}{6/\sqrt{3} } = 2.31$

P(Mean amount of 3 proffessors works for more than 60 hours)

= P( $\bar{x}$ >60)

= P(Z > 2.31)

  = 1- P(Z < 2.31)

= 1 - 0.9896

= 0.0104