Question

Ch. 6 #6: please assist with answers for a, b, and c in percentages:

The average rent in a city is $1,680 per month with a standard deviation of $200. Assume rent follows the normal distribution. [You may find it useful to reference the z table.] a. What percentage of rents are between $1,480 and $1,880? (Round your answer to the nearest whole percent.) b. What percentage of rents are less than $1,480? (Round your answer to 1 decimal place.) c. What percentage of rents are greater than $2,080? (Round your answer to 1 decimal place.)

a. What percentage of rents are between $1,480 and $1,880? (Round your answer to the nearest whole percent.)

Percentage of rents:________%

b. What percentage of rents are less than $1,480? (Round your answer to 1 decimal place.)

Percentage of rents:________%

c. What percentage of rents are greater than $2,080? (Round your answer to 1 decimal place.)

Percentage of rents:________%

Answer 1

Solution :

Given that,

mean = $\mu$ = $1,680

standard deviation = $\sigma$ =$200.

A) P ( 1,480 < x < 1,880)

P ( 1,480 - 1,680 / 200) < ( x -  $\mu$/ $\sigma$) < ( 1,880 - 1,680 / 200)

P ( - 200 / 200< z < 200/ 200)

P (- 1 < z < 1 )

P (z < 1 ) - p ( z < - 1 )

Using z table

= 0.8413 - 0.1587

= 0.6826

Probability = 68,3%

B) P ( x < 1,480)

P( x -  $\mu$/ $\sigma$) < ( 1,480 - 1,680  / 200)

P ( z < - 200/ 200)

P ( z < - 1 )

Using z table

= 0.1587

Probability =15.9%

C) P (x < 2,080)

P ( x -  $\mu$/ $\sigma$) < ( 2,080 - 1,680  / 200)

P ( z < 400/ 200)

P ( z < 2)

Using z table

= 0.9772

Probability = 97.7%