The distribution of durations for which apartments remain empty after the resident moves out for one property management company over the past 101010 years was approximately normal with mean \mu = 85μ=85mu, equals, 85 days and standard deviation \sigma = 29σ=29sigma, equals, 29 days. The property management company intends to update the kitchen appliances in the apartments that were empty for top 10\%10%10, percent of durations. 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.

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Answer:

The minimum duration for which an apartment remained empty for the company to update the kitchen appliances is 122 days.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

[tex]\mu = 85, \sigma = 29[/tex]

What is the minimum duration for which an apartment remained empty for the company to update the kitchen appliances?

Top 10%, which is the 100-10 = 90th percentile.

The 90th percentile is X when Z has a pvalue of 0.9, so X when Z = 1.28. Then

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.28 = \frac{X - 85}{29}[/tex]

[tex]X - 85 = 1.28*29[/tex]

[tex]X = 122.12[/tex]

Rounded to the nearest whole number,

The minimum duration for which an apartment remained empty for the company to update the kitchen appliances is 122 days.

Answer:

123 days

Step-by-step explanation:

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