Question

you have information to suggest that a certain continuous variable of a population has a mean of \\( \mu = 13.24 \\) and a standard deviation of \\( \sigma = 5.12 \\). you are to randomly pick \\( n = 54 \\) individuals from this population and observe the value of the population variable on each. this value is to measured as \\( x \\).
after the random sample of \\( n = 54 \\) has been taken, you are asked to consider the behavior of the statistic \\( \overline{x} \\).
(a) complete the statement below. use as many decimals as you can.
the distribution of \\( \overline{x} \\) is approximately normal with a mean \\( \mu_{\overline{x}} = 13.24 \\) and a standard deviation \\( \sigma_{\overline{x}} = 0.6916 \\)
(b) find the probability that mean of the sample of \\( n = 54 \\) is between 12.64 and 13.74. use at least four decimals in your z - values, and enter your answer to at least four decimals.

Explanation:

Step1: Calculate the z - scores

The formula for the z - score is \(z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}\).
We know that \(\mu_{\bar{x}} = 13.24\), \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{5.12}{\sqrt{54}}\approx0.6974\) (Note: There was a mistake in the given \(\sigma_{\bar{x}}\) value in part (a), the correct formula is used here).
For \(\bar{x}_1 = 12.64\):
\(z_1=\frac{12.64 - 13.24}{0.6974}=\frac{- 0.6}{0.6974}\approx - 0.86\)
For \(\bar{x}_2 = 13.74\):
\(z_2=\frac{13.74 - 13.24}{0.6974}=\frac{0.5}{0.6974}\approx0.72\)

Step2: Find the probabilities using the standard normal distribution

We know that \(P(a<Z < b)=\Phi(b)-\Phi(a)\), where \(\Phi(z)\) is the cumulative distribution function of the standard normal distribution.
From the standard - normal table, \(\Phi(-0.86)=0.1949\) and \(\Phi(0.72)=0.7642\)

Step3: Calculate the required probability

\(P(12.64<\bar{X}<13.74)=\Phi(0.72)-\Phi(- 0.86)\)
\(P(12.64<\bar{X}<13.74)=0.7642-0.1949 = 0.5693\)

Answer:

\(0.5693\)