Question

A study was conducted by a research center. It reported that most shoppers have a specific spending limit in place while shopping online. The reports indicate that men spend an average of $240 online before they decide to visit a store. If the spending limit is normally distributed and the standard deviation is $20.
A. Find the probability that a male spent less than $210 online before deciding to visit a store.
B. Find the probability that a male spent between $270 and $300 online before deciding to visit a store.
C. Ninety percent of the amounts spent online by a male before deciding to visit a store are less than what value?

Answer 1

Answer:

(A) The probability that a male spent less than $210 online before deciding to visit a store is 0.0668.

(B) The probability that a male spent between $270 and $300 online before deciding to visit a store is 0.0655.

(C) Ninety percent of the amounts spent online by a male before deciding to visit a store is less than $265.632.

Step-by-step explanation:

We are given that the reports indicate that men spend an average of $240 online before they decide to visit a store. If the spending limit is normally distributed and the standard deviation is $20.

Let X = the spending limit

The z-score probability distribution for the normal distribution is given by;

                           Z  =  $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = mean spending limit = $240

           $\sigma$ = standard deviation = $20

So, X ~ Normal($\mu=\ 240,\sigma^{2} =\ 20^{2}$)

(A) The probability that a male spent less than $210 online before deciding to visit a store is given by = P(X < $210)

     P(X < $210) = P( $\frac{X-\mu}{\sigma}$ < $\frac{\ 210-\ 240}{\ 20}$ ) = P(Z < -1.50) = 1 - P(Z $\leq$ 1.50)

                                                            = 1 - 0.9332 = 0.0668

The above probability is calculated by looking at the value of x = 1.50 in the z table which has an area of 0.9332.

(B) The probability that a male spent between $270 and $300 online before deciding to visit a store is given by = P($270 < X < $300)

     P($270 < X < $300) = P(X < $300) - P(X $\leq$ $270)

     P(X < $300) = P( $\frac{X-\mu}{\sigma}$ < $\frac{\ 300-\ 240}{\ 20}$ ) = P(Z < 3) = 0.9987

     P(X $\leq$ $270) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{\ 270-\ 240}{\ 20}$ ) = P(Z $\leq$ 1.50) = 0.9332

The above probability is calculated by looking at the value of x = 3 and x = 1.50 in the z table which has an area of 0.9987 and 0.9332 respectively.

Therefore, P($270 < X < $300) = 0.9987 - 0.9332 = 0.0655.

(C) Now, we have to find ninety percent of the amounts spent online by a male before deciding to visit a store is less than what value, that is;

         P(X < x) = 0.90     {where x is the required value}

         P( $\frac{X-\mu}{\sigma}$ < $\frac{x-\ 240}{\ 20}$ ) = 0.90

         P(Z < $\frac{x-\ 240}{\ 20}$ ) = 0.90

In the z table, the critical value of z that represents the bottom 90% of the area is given as 1.2816, i.e;

                     $\frac{x-\ 240}{\ 20}=1.2816$

                     $x-240=1.2816\times 20$

                     $x=240 + 25.632$

                     x = 265.632

Hence, Ninety percent of the amounts spent online by a male before deciding to visit a store is less than $265.632.