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) $1,010 or more?

(b) Between $880 and $1,120?

(c) Between $825 and $935?

(d) Less than $730?

Answer 1

(a)

= 951

= 96

To find P(X1010):

Z = (1010 = 951)/96

= 0.61

By Technology, cumulative area under standard normal curve = 0.7291

So,

P(X1010) = 1 - 0.7291 = 0.2709

So,

Answer is:

0.2709

(b)

= 951

= 96

To find P(880 < X < 1120):

For X = 880::

Z = (880 = 951)/96

= - 0.74

By Technology, cumulative area under standard normal curve = 0.2297

For X = 1120:

Z = (1120 = 951)/96

= 1.76

By Technology, cumulative area under standard normal curve = 0.9608

So,

P(880 < X < 1120) = 0.9608 - 0.2297 = 0.7311

So,

Answer is:

0.7311

(c)

= 951

= 96

To find P(825 < X < 935):

For X = 825::

Z = (825 = 951)/96

= - 1.31

By Technology, cumulative area under standard normal curve = 0.0951

For X = 935:

Z = (935 = 951)/96

= - 0.17

By Technology, cumulative area under standard normal curve = 0.4325

So,

P(825 < X < 935):= 0.4325 - 0.0951 = 0.3374

So,

Answer is:

0.3374

(d)

= 951

= 96

To find P(X < 730):

Z = (730 = 951)/96

= - 2.30

By Technology, cumulative area under standard normal curve = 0.0107

So,

P(X < 730):= 0.0107

So,

Answer is:

0.0107