Question

The average cost of a one-bedroom apartment in a town is $650 per month. What is the probability of randomly selecting a sample of 50 one-bedroom apartments in this town and getting a sample mean of less than $630 if the population standard deviation is $100?

Answer 1

Answer:

Probability of randomly selecting a sample of 50 one-bedroom apartments in this town and getting a sample mean of less than $630 is 0.0793.

Step-by-step explanation:

We are given that the average cost of a one-bedroom apartment in a town is $650 per month. Assume the population standard deviation is $100.

Also, a sample of 50 one-bedroom apartments is selected.

Let $\bar X$ = sample mean cost

The z-score probability distribution for sample mean is given by ;

                Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population average cost = $650

            $\sigma$ = population standard deviation = $100

            n = sample of one-bedroom apartments = 50

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.  

Now, the probability that the sample mean is less than $630 is given by = P($\bar X$ < $630)

   P($\bar X$ < $630) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{630 - 650}{\frac{100}{\sqrt{50} } }$ ) = P(Z < -1.41) = 1 - P(Z $\leq$ 1.41)

                                                        = 1 - 0.9207 = 0.0793

The above probability is calculated using z table by looking at value of x = 1.41 in the z table which have an area of 0.92073.

Therefore, probability that the sample mean is less than $630 is 0.0793.

Answer 2

We can use standard normal variate and use z tables to find the needed probability.

The needed probability is 0.0793

Given that:

• The average cost of one bedroom apartment = $\mu = \ 650$
• The standard deviation of population  = $\sigma = \ 100$
• The size of the sample = $n = 50$
• The mean of sample is needed to be less than $630

Let the sample mean be tracked by random variable $\overline{X}$

Transforming $\overline{X}$ to standard normal variate

Since standard deviation is known already, we can transform the r.v. approximately into standard normal variate by:

$Z = \dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

Then we will have:

$P(\overline{X} < 630) = P(Z < \dfrac{630 - 650}{\dfrac{100}{\sqrt{50}}} ) = P(Z < -\sqrt{2})\\\\P(\overline{X} < 630) = P(Z < -1.41) = 1 - P(Z \leq 1.41) = 1 - 0.9207 = 0.0793$

The value 0.9207 is p value for z-score 1.41 seen from the z-score table.

Thus, the needed probability is 0.0793

Learn more about z-score here:

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