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

Answer:

(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.

Step-by-step 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  =  $\frac{X-\mu}{\sigma}$  ~ 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( $\frac{X-\mu}{\sigma}$ < $\frac{\ 1000-\ 6312}{\sigma}$ ) = 0.05

                      P(Z < $\frac{\ 1000-\ 6312}{\sigma}$ ) = 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;

                           $\frac{\ 1000-\ 6312}{\sigma}=-1.645$                

                            $\sigma=\frac{-\ 5312}{-1.645}$  = 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( $\frac{X-\mu}{\sigma}$ < $\frac{\ 6000-\ 6312}{\ 3229.18}$ ) = P(Z < -0.09) = 1 - P(Z $\leq$ 0.09)

                                                            = 1 - 0.5359 = 0.4641

 P(X $\leq$ $4000) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{\ 4000-\ 6312}{\ 3229.18}$ ) = P(Z $\leq$ -0.72) = 1 - P(Z < 0.72)

                                                            = 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( $\frac{X-\mu}{\sigma}$ > $\frac{x-\ 6312}{3229.18}$ ) = 0.03

                    P(Z > $\frac{x-\ 6312}{3229.18}$ ) = 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;

                           $\frac{x-\ 6312}{3229.18}=1.88$                

                         ${x-\ 6312}=1.88\times 3229.18$  

                          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.