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

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.

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

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step-by-step explanation:

step-by-step explanation:

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