Question

Historical data show that customers who download music from a popular Web service spend approximately ​$20 per​ month, with a standard deviation of ​$3. Assume the spending follows the normal probability distribution. Find the probability that a customer will spend at least ​$17 per month. How much​ (or more) do the top 6​% of customers​ spend?

What is the probability that a customer will spend at least $17 per month?

How much do the top 6% of customers spend?

Answer 1

Solution :

Given that,

mean = $\mu$ = 20

standard deviation = $\sigma$ = 3

P(x >17 ) = 1 - P(x< 17)

= 1 - P[(x -$\mu$) / $\sigma$ < (17-20) /3 ]

= 1 - P(z < -1)

Using z table

= 1 - 0.1587

probability= 0.8413

B.

P(Z > z) = 6%

= 1 - P(Z < z) = 0.06  

= P(Z < z) = 1 - 0.06

= P(Z < z ) = 0.94

z =1.56

Using z-score formula,

x = z * $\sigma$ + $\mu$

=1.56*3+20

=24.68