Question

The average rent in a city is $1,410 per month with a standard deviation of $290. Assume rent follows the normal distribution. [You may find it useful to reference the z table.] a. What percentage of rents are between $830 and $1,990? (Round your answer to the nearest whole percent.) b. What percentage of rents are less than $830? (Round your answer to 1 decimal place.) c. What percentage of rents are greater than $2,280? (Round your answer to 1 decimal place.)

Answer 1

Solution :

Given that mean μ = 1410 , standrad deviation σ = 290

a. => P(830 < x < 1990) = P((830 - 1410)/290 < (x - μ)/σ < (1990 - 1410)/290)

= P(-2 < Z < 2)

= 0.9544

= 95.44%

= 95.4% (rounded)

b. => P(x < 830) = P((x - μ)/σ < (830 - 1410)/290)

= P(Z < -2)

= 1 − P(Z < 2)

= 1 − 0.9772

= 0.0228

= 02.28%

= 2.3% (rounded)

c. => P(x > 2280) = P((x - μ)/σ > (2280 - 1410)/290)

= P(Z > 3)

= 1 − P(Z < 3)

= 1 − 0.9987

= 0.0013

= 0.13%

= 0.1% (rounded)