Question

USING EXCEL The XO Group Inc. conducted a survey of 13,000 brides and grooms married in the United States and found that the average cost of a wedding is $29,858 (XO Group website, January 5, 2015). Assume that the cost of a wedding is normally distributed with a mean of $29,858 and a standard deviation of $5600. What is the probability that a wedding costs less than $20,000? What is the probability that a wedding costs between $20,000 and $30,000? For a wedding to be among the 5% most expensive, how much would it have to cost?

Answer 1

Solution:

Given:  the cost of a wedding is normally distributed with a mean of $29,858 and a standard deviation of $5600.

That is:

We have to use Excel to find following probabilities:

Part a) What is the probability that a wedding costs less than $20,000?

P( X < 20000) = ............?

=NORM.DIST( x , mean , sd , cumulative)

=NORM.DIST(20000,29858,5600,TRUE)

=0.0391736

=0.0392

Thus

P( X < 20000) = 0.0392

Part b)  What is the probability that a wedding costs between $20,000 and $30,000?

P( 20000 < X < 30000) = ..............?

P( 20000 < X < 30000) = P( X < 30000) - P( X < 20000)

Use following Excel command:

=NORM.DIST( Upper x , mean , sd , cumulative) - NORM.DIST( Lower x , mean , sd , cumulative)

=NORM.DIST( 30000 , 29858 , 5600 ,TRUE)-NORM.DIST( 20000 , 29858 , 5600 ,TRUE)

=0.47094132

=0.4709

That is:

P( 20000 < X < 30000) = 0.4709

Part c) For a wedding to be among the 5% most expensive, how much would it have to cost?

That is find x value such that: P( X > x ) = 5%

then P( X < x ) = 100 - 5 = 95%

thus use following excel command:

=NORM.INV( probability , Mean, SD )

=NORM.INV(0.95,29858,5600)

=39069.1803

=39069.18

Thus x = $39,069.18