Question

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.

Answer 1

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 85, \sigma = 29$

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

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 85}{29}$

$X - 85 = 1.28*29$

$X = 122.12$

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 2

Answer:

123 days

Step-by-step explanation: