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?

(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  =    ~ N(0,1)

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

            = standard deviation

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

                      P(X < $1,000) = 0.05

                      P( < ) = 0.05

                      P(Z < ) = 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;

                             = 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 $4000)

 P(X < $6000) = P( < ) = P(Z < -0.09) = 1 - P(Z 0.09)

                                                            = 1 - 0.5359 = 0.4641

 P(X $4000) = P( ) = P(Z -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( > ) = 0.03

                    P(Z > ) = 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 + 6070.86 = $12382.86

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

the cat is blue because the probability of extension towards the prefix of the diameter of the circle could only mathematically equal to the sum of the roots of the carrots from   your garden.

step-by-step explanation:

i believe the answer is the polygon, heptagon.