Question

Suppose that the distribution for total amounts spent by students vacationing for a week in Florida is normally distributed with a mean of 650 and a standard deviation of 120. Suppose you take a SRS of 30 students from this distribution. What is the probability that an SRS of 30 students will spend an average of between 600and700? Round to five decimal places.

Answer 1

Let x be the  total amounts spent by students .

X follows normal distribution with mean $\mu$ = 650 and standard deviation $\sigma$ = 120

We take simple random sample of size n = 30

We are asked to find average spend of 30 students ( $\bar{x}$ ) is between 600 and 700

We have to find P( 600 <= $\bar{x}$ <= 700)

According to sampling distribution of the sample mean $\bar{x}$ , it follows approximately normal distribution with mean $\mu _{\bar{x}}$ = $\mu$ and standard deviation   $\sigma _{\bar{x}}$ = $\frac{\sigma }{\sqrt{n}}$

Therefore here mean of $\bar{x}$ ,( $\mu _{\bar{x}}$ ) = 650 and

standard deviation of $\bar{x}$ , $\sigma _{\bar{x}}$ = $\frac{120 }{\sqrt{30}}$ = 21.9089

We can use TI-84 calculator to find this probability.

Press 2ND --->VARS ---->Select normalcdf() function and hit enter.

Then plug the given values as follows :

Scroll to paste and hit enter 2 times ,

Final answer : 0.97752 rounded to 5 decimal places .