Question

The distribution of durations for which apartments remain empty after the resident moves out for one property management company over the past 10 1010 years was approximately normal with mean μ = 85 μ=85mu, equals, 85 days and standard deviation σ = 29 σ=29sigma, equals, 29 days. The property management company tags the files of the apartments that were empty for the shortest 5 % 5%5, percent of durations to have priority cleaning the next time their residents move out

Answer 1

A part of the question is missing which is;

What is the minimum duration for which an apartment remained empty for the company to update the kitchen appliances? Round to the nearest whole number.

Answer:

133 days

Step-by-step explanation:

We are given;

Population mean; μ = 85 days Population standard deviation; σ = 29 days

significance level; α = 5% = 0.05

Critical value of z at significance level of 0.05 = 1.645

Now, formula for z-score is;

z = (x - μ)/σ

Where x is the minimum duration for which an apartment remained empty for the company to update the kitchen appliances.

1.645 = (x - 85)/29

(1.645 × 29) = x - 85

47.705 = x - 85

x = 85 + 47.705

x = 132.705

x ≈ 133 days