Question

A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimate m.a.What is the probability that the sample mean will be within ±5 of the population mean?b.What is the probability that the sample mean will be within ±10 of the population mean?

Answer 1

Answer:

a) 68% probability that the sample mean will be within ±5 of the population mean.

b) 95% probability that the sample mean will be within ±10 of the population mean.

Step-by-step explanation:

To solve this problem, we have to understand the Empirical Rule and the Central Limit Theorem

Empirical Rule

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s= \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

Mean: $\mu = 200$

Standard deviation of the population: $\sigma = 50$

Size of the sample: $n = 100$

Standard deviation for the sample mean: $s = \frac{50}{\sqrt{100}} = 5$

a.What is the probability that the sample mean will be within ±5 of the population mean?

5 is one standard deviation from the mean.

By the Empirical Rule, 68% probability that the sample mean will be within ±5 of the population mean.

b.What is the probability that the sample mean will be within ±10 of the population mean?

10 is two standard deviations from the mean

By the Empirical Rule, 95% probability that the sample mean will be within ±10 of the population mean.