Question

consider the sampling distribution of x for random samples of 65 customers satisfactons ratings.Determine the probability of observing a sample mean greater than or equal to 42.95 when we assume that mean equals 42.

Answer 1

Answer on Question #70350 – Math – Statistics and Probability

Question

Consider the sampling distribution of $x$ for random samples of 65 customer satisfactions ratings. Determine the probability of observing a sample mean greater than or equal to 42.95 when we assume that mean equals 42.

Solution

Assume that $\sigma$, standard deviation equals 2.64.

$$\begin{array}{l} \mu = 42, \sigma = 2.64, n = 65, \overline{x} = 42.95 \\ \mu_{\overline{x}} = 42, \sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}} = \frac{2.64}{\sqrt{65}} \\ P(\overline{x} \geq 42.95) = P\left(z \geq \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\right) = P\left(z \geq \frac{42.95 - 42}{\frac{2.64}{\sqrt{65}}}\right) = P(z \geq 2.9012) = \\ = 0.0019 \\ \end{array}$$

Answer provided by https://www.AssignmentExpert.com