According to the June 2004 Readers Digest article Only in Am

According to the June 2004 Readers\' Digest article \"Only in America,\" the average amount that a 17-year-old spends on his or her high school prom is $640 Assume that the amounts spent are normally distributed with a standard deviation of $185. (Give your answers correct to four decimal places.)

(a) Find the probability that the mean cost to attend a high school prom for 26 randomly selected high school 17-year-olds is between $548 and $704.

(b) Find the probability that the mean cost to attend a high school prom for 26 randomly selected high school 17-year-olds is greater than $749.

Solution

Let X be the random variable that  the amounts spent on a 17-year-old spends on his or her high school prom.

X ~ Normal ( µ = $640, = $185)

(a) Find the probability that the mean cost to attend a high school prom for 26 randomly selected high school 17-year-olds is between $548 and $704.

n = 26

We have to find the P(548 < Xbar < 704)

where Xbar is sample mean.

First we have to convert Xbar into standard normal distribution.

We know than Xbar ~ Normal(mean = µ, sd = /sqrt(n) )

sd =  /sqrt(n) = 185 / sqrt(26) = 36.2815

Z-score for Xbar = 548 is,

Z-score = (Xbar - mean) / sd

Z-score = (548-640) / 36.2815 = -2.5357

Z-score for Xbar = 704 is,

Z-score = (704 - 640) / 36.2815 = 1.7640

So we have to find P(-2.5357 < Z < 1.7640) = P(Z < 1.7640) - P(Z < -2.5357)

This probability we can find by using EXCEL.

syntax :

=NORMSDIST(z)

where, z is the test statistic value.

P(-2.5357 < Z < 1.7640) = 0.9611 - 0.0056 = 0.9555

(b) Find the probability that the mean cost to attend a high school prom for 26 randomly selected high school 17-year-olds is greater than $749.

Here we have to find P(Xbar > 749)

Convert Xbar into Z-score.

Z-score for Xbar = 749 is,

Z-score = (749 - 640) / 36.2815

Z-score = 3.0043

P(Z >= 3.0043) = 1 - P(Z < 3.0043)

P(Z >= 3.0043) = 1 - 0.9987 = 0.0013