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

(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.

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 =
`(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(
`(X-\mu)/(\sigma)` <
`(\$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(
`(X-\mu)/(\sigma)` <
`(\$300-\$240)/(\$20)` ) = P(Z < 3) = 0.9987

P(X
`\leq` $270) = P(
`(X-\mu)/(\sigma)`
`\leq`
`(\$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(
`(X-\mu)/(\sigma)` <
`(x-\$240)/(\$20)` ) = 0.90

P(Z <
`(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;

`(x-\$240)/(\$20)=1.2816`

`x-240=1.2816* 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.