Question

M/PF Research, Inc. lists the average monthly apartment rent in some of the most expensive apartment rental locations in the United States. According to their report, the average cost of renting an apartment in Minneapolis is $951. Suppose that the standard deviation of the cost of renting an apartment in Minneapolis is $96 and that apartment rents in Minneapolis are normally distributed. If a Minneapolis apartment is randomly selected, what is the probability that the price is: (Round the values of z to 2 decimal places. Round answers to 4 decimal places.) (a) $1,020 or more? (b) Between $880 and $1,130? (c) Between $830 and $940? (d) Less than $770?

Answer 1

Answer:

(a) $1,020 or more = 0.2358

(b) Between $880 and $1,130 = 0.7389

(c) Between $830 and $940 = 0.3524

(d) Less than $770 = 0.0294

Step-by-step explanation:

We are given that According to M/PF Research, Inc. report, the average cost of renting an apartment in Minneapolis is $951.

Suppose that the standard deviation of the cost of renting an apartment in Minneapolis is $96 and that apartment rents in Minneapolis are normally distributed.

Let X = apartment rents in Minneapolis

So, X ~ Normal($\mu= 951,\sigma^{2} = 96^{2}$)

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

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

where,  $\mu$ = average cost of renting an apartment = $951

            $\sigma$ = standard deviation = $96

(a) Probability that the price is $1,020 or more is given by = P(X $\geq$ $1,020)

    P(X $\geq$ $1,020) = P( $\frac{X-\mu}{\sigma}$ $\geq$ $\frac{1,020-951}{96}$ ) = P(Z $\geq$ 0.72) = 1 - P(Z < 0.72)

                                                               = 1 - 0.76424 = 0.2358

The above probability is calculated by looking at the value of x = 0.72 in the z table which gives an area of 0.76424.

(b) Probability that the price is between $880 and $1,130 is given by = P($880 < X < $1,130) = P(X < $1,130) - P(X $\leq$ 880)

    P(X < $1,130) = P( $\frac{X-\mu}{\sigma}$ < $\frac{1,130-951}{96}$ ) = P(Z < 1.86) = 0.96856

     P(X $\leq$ $880) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{880-951}{96}$ ) = P(Z $\leq$ -0.74) = 1 - P(Z < 0.74)

                                                         = 1 - 0.77035 = 0.22965

The above probability is calculated by looking at the value of x = 1.86 and x = 0.74 in the z table which gives an area of 0.96856 and 0.77035 respectively.

Therefore, P($880 < X < $1,130) = 0.96856 - 0.22965 = 0.7389

(c) Probability that the price is between $830 and $940 is given by = P($830 < X < $940) = P(X < $940) - P(X $\leq$ 830)

    P(X < $940) = P( $\frac{X-\mu}{\sigma}$ < $\frac{940-951}{96}$ ) = P(Z < -0.11) = 1 - P(Z $\leq$ 0.11)

                                                          = 1 - 0.5438 = 0.4562

     P(X $\leq$ $830) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{830-951}{96}$ ) = P(Z $\leq$ -1.26) = 1 - P(Z < 1.26)

                                                         = 1 - 0.89617 = 0.10383

The above probability is calculated by looking at the value of x = 0.11 and x = 1.26 in the z table which gives an area of 0.5438 and 0.89617 respectively.

Therefore, P($830 < X < $940) = 0.4562 - 0.10383 = 0.3524

(d) Probability that the price is Less than $770 is given by = P(X < $770)

    P(X < $770) = P( $\frac{X-\mu}{\sigma}$ < $\frac{770-951}{96}$ ) = P(Z < -1.89) = 1 - P(Z $\leq$ 1.89)

                                                         = 1 - 0.97062 = 0.0294

The above probability is calculated by looking at the value of x = 1.89 in the z table which gives an area of 0.97062.