Question

A simple random sample of size n = 49 is obtained from a population with u = 89 and o = 21.

Answer 1

Answer:

a) B. The distribution is approximately normal.

b) 0.0322 = 3.22%

c) 0.0202 = 2.02%

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean  and standard deviation , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean and standard deviation:

$\mu = 89, \sigma = 21$

Sample of 49:

This means that $n = 49, s = \frac{21}{\sqrt{49}} = 3$

(a) Describe the sampling distribution of x.

By the Central Limit Theorem, approximately normal, and the correct answer is given by option B.

(b) What is P (x > 94.55) ?

This is 1 subtracted by the p-value of Z when X = 94.55, so:

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

By the Central Limit Theorem

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

$Z = \frac{94.55 - 89}{3}$

$Z = 1.85$

$Z = 1.85$ has a p-value of 0.9678.

1 - 0.9678 = 0.0322

So 0.0322 = 3.22%

Question c:

This is the p-value of Z when X = 82.85. So

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

$Z = \frac{82.85 - 89}{3}$

$Z = -2.05$

$Z = -2.05$ has a p-value of 0.0202.

So

0.0202 = 2.02%