Question

1. the mean rent paid by all tenants in a small city is $1550 with a standard deviation of $225. however, the population distribution of rents for all tenants in this city is skewed to the right. a) describe the shape of the sampling distribution of the mean rent paid when a random sample of size 100 is chosen. b) calculate the mean and standard deviation of the mean rent paid of these 100 tenants. 5. the edwards’s theater chain has studied its movie customers to determine how much money they spend on concessions. the study revealed that the spending distribution is approximately normally distributed with a mean of $5.15 and a standard deviation of $1.37. if a random sample of 20 customers are selected, a) what is the probability that they will spend, on average, less than $5.00 on concessions? b) what is the average spending amount that corresponds to the highest 25% of these customers?

Explanation:

4a)

Step1: Apply central - limit theorem

According to the central - limit theorem, when the sample size $n = 100$ (where $n\geq30$), the sampling distribution of the sample mean is approximately normal regardless of the shape of the population distribution.

4b)

Step1: Find the mean of the sampling distribution

The mean of the sampling distribution of the sample mean $\mu_{\bar{x}}$ is equal to the population mean $\mu$. So, $\mu_{\bar{x}}=\mu = 1550$.

Step2: Find the standard deviation of the sampling distribution

The standard deviation of the sampling distribution of the sample mean (also known as the standard error) is given by $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma = 225$ and $n = 100$. Then $\sigma_{\bar{x}}=\frac{225}{\sqrt{100}}=\frac{225}{10}=22.5$.

5a)

Step1: Calculate the z - score

The formula for the z - score of the sample mean is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$, where $\mu = 5.15$, $\sigma = 1.37$, $n = 20$, and $\bar{x}=5.00$. So, $z=\frac{5 - 5.15}{\frac{1.37}{\sqrt{20}}}=\frac{- 0.15}{\frac{1.37}{4.472}}\approx\frac{-0.15}{0.306}\approx - 0.49$.

Step2: Find the probability

We want $P(\bar{X}<5.00)$, which is equivalent to $P(Z < - 0.49)$. Looking up in the standard - normal table, $P(Z < - 0.49)=0.3121$.

5b)

Step1: Find the z - score corresponding to the upper 25%

The z - score corresponding to the upper 25% (or the 75th percentile) is $z = 0.674$ (from the standard - normal table).

Step2: Solve for $\bar{x}$

Using the z - score formula $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$, we can solve for $\bar{x}$. Rearranging gives $\bar{x}=\mu+z\frac{\sigma}{\sqrt{n}}$. Substituting $\mu = 5.15$, $z = 0.674$, $\sigma = 1.37$, and $n = 20$, we have $\bar{x}=5.15+0.674\times\frac{1.37}{\sqrt{20}}=5.15+0.674\times0.306=5.15 + 0.206=5.356\approx5.36$.

Answer:

4a) Approximately normal.
4b) Mean: $1550$, Standard deviation: $22.5$
5a) $0.3121$
5b) $\$5.36$