Question

A purveyor of designer maternity wear sells dresses and pants priced around ​$250 each for an average total sale of ​$1,200. The total sale has a normal distribution with a standard deviation of ​$450. Complete parts a through c below. a. Calculate the probability that a randomly selected customer will have a total sale of more than ​$1,700. The probability is 0.1314. ​(Round to four decimal places as​ needed.) b. Compute the probability that the total sale will be within plus or minus2 standard deviations of the mean total sales.

Answer 1

Let , X be the total sale.

X follows normal distribution with = $1200 and standard deviation = $450

a )

We have to find P( x > 1700 )

P( x > 1700 ) = 1 - P( x <= 1700 )

Using Excel, =NORMDIST( x, , , 1 )

P( x <= 1700 ) = NORMDIST( 1700 , 1200, 450 , 1 ) = 0.86674

So, P( x > 1700 ) = 1 - 0.86674 = 0.1333

Answer :

Probability that a randomly selected customer will have a total sale of more than ​$1,700 is 0.1333

b)

We have to find P( 1200 - 2*450 < x <1200 + 2*450 )

That is P( 300 < x < 2100 )

P( 300 < x < 2100 ) = P( x < 2100) - P( x < 300 )

Using Excel, =NORMDIST( x, , , 1 )

P( x < 2100) = NORMDIST( 2100 , 1200, 450 , 1 ) = 0.97725

P( x < 300 ) = NORMDIST( 300 , 1200, 450 , 1 ) = 0.02275

So, P( 300 < x < 2100 ) = 0.97725 - 0.02275 = 0.9545

Answer :

Probability that the total sale will be within plus or minus2 standard deviations of the mean total sales is 0.9545