Question

The average annual amount American households spend for daily transportation is $6312 (Money, August 2001). Assume that the amount spent is normally distributed.a. Suppose you learn that 5% of American households spend less than $1000 for dailytransportation. What is the standard deviation of the amount spent?b. What is the probability that a household spends between $4000 and $6000?c. What is the range of spending for the 3% of households with the highest daily transportationcost?

Answer 1

(a) The standard deviation of the amount spent is $3229.18.

(b) The probability that a household spends between $4000 and $6000 is 0.2283.

(c) The range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.

Explanation:

We are given that the average annual amount American households spend on daily transportation is $6312 (Money, August 2001). Assume that the amount spent is normally distributed.

(a) It is stated that 5% of American households spend less than $1000 for daily transportation.

Let X = the amount spent on daily transportation

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

Z =
`(X-\mu)/(\sigma)` ~ N(0,1)

where,
`\mu` = average annual amount American households spend on daily transportation = $6,312

`\sigma` = standard deviation

Now, 5% of American households spend less than $1000 on daily transportation means that;

P(X < $1,000) = 0.05

P(
`(X-\mu)/(\sigma)` <
`(\$1000-\$6312)/(\sigma)` ) = 0.05

P(Z <
`(\$1000-\$6312)/(\sigma)` ) = 0.05

In the z-table, the critical value of z which represents the area of below 5% is given as -1.645, this means;

`(\$1000-\$6312)/(\sigma)=-1.645`

`\sigma=(-\$5312)/(-1.645)` = 3229.18

So, the standard deviation of the amount spent is $3229.18.

(b) The probability that a household spends between $4000 and $6000 is given by = P($4000 < X < $6000)

P($4000 < X < $6000) = P(X < $6000) - P(X
`\leq` $4000)

P(X < $6000) = P(
`(X-\mu)/(\sigma)` <
`(\$6000-\$6312)/(\$3229.18)` ) = P(Z < -0.09) = 1 - P(Z
`\leq` 0.09)

= 1 - 0.5359 = 0.4641

P(X
`\leq` $4000) = P(
`(X-\mu)/(\sigma)`

= 1 - 0.7642 = 0.2358

Therefore, P($4000 < X < $6000) = 0.4641 - 0.2358 = 0.2283.

(c) The range of spending for 3% of households with the highest daily transportation cost is given by;

P(X > x) = 0.03 {where x is the required range}

P(
`(X-\mu)/(\sigma)` >
`(x-\$6312)/(3229.18)` ) = 0.03

P(Z >
`(x-\$6312)/(3229.18)` ) = 0.03

In the z-table, the critical value of z which represents the area of top 3% is given as 1.88, this means;

`(x-\$6312)/(3229.18)=1.88`

`{x-\$6312}=1.88* 3229.18`

x = $6312 + 6070.86 = $12382.86

So, the range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.