Question

Grandparents spent an average of $450 on their grandkids in 2019. Assume that the amount spent on grandkids is normally distributed and that the standard deviation is $45. (a) What is the probability that a randomly selected grandparent spent more than $475? (Round Z-value to two decimal places and probability to four decimal places.) (b) What is the probability that a randomly selected grandparent spent between $415 and $500? (Round Z-value to two decimal places and probability to four decimal places.)

Answer 1

Solution :

Given ,

mean = = $450

standard deviation = = $45

P(x > $475) = 1 - P(x<475 )

= 1 - P[(x -) / < (475-450) /45 ]

= 1 - P(z <0.56 )

Using z table

= 1 - 0.7123

= 0.2877

probability= 0.2877

(b)

P($415< x <$500 ) = P[(415-450) / 45< (x - ) / < (500-450) /45 )]

= P( -0.78< Z <1.11 )

= P(Z <1.11 ) - P(Z < -0.78)

Using z table   

= 0.8665-0.2177

probability= 0.6488