Question

The population mean and standard deviation are given below. Find the required probability and determine whether the given sample mean would be consideredunusualFor a sample of n=61, find the probability of a sample mean being less than 19.6 if u = 20 and 0 = 1.18.Click the icon to view page 1 of the standard normal table.Click the icon to view page 2 of the standard normal table.For a sample of n=61, the probability of a sample mean being less than 19.6 if u = 20 and = 1.18 is I(Round to four decimal places as needed.)Would the given sample mean be considered unusual?The sample mean be considered unusual because it has a probability that is than 5%

Answer 1

First we will obtain the value for the Zscore using the formula

`\begin{gathered} \text{Zscore = }\frac{(x-\mu)/(\sigma)}{\sqrt[]{n}} \\ \\ \text{Where }x\text{ = }sample\text{ mean = 19.6} \\ \mu=\text{ population mean = }20 \\ \sigma=\text{ standard deviation = }1.18 \\ n=\text{ }sample\text{ size = 61} \end{gathered}`

This becomes

`\begin{gathered} \text{Zscore = }\frac{(x-\mu)/(\sigma)}{\sqrt[]{n}} \\ \\ \text{Zscore = }\frac{(19.6-20)/(1.18)}{\sqrt[]{61}} \\ \\ \text{Zscore = }\frac{-0.338983}{\sqrt[]{61}} \\ \\ \text{Zscore = -0.0434} \end{gathered}`

Now using the Zscore calculator, we obtain the score for -0.0434

This becomes

[tex]P(xTherefore, the probability = 0.4827 to four decimal places.

Would the sample mean be considered unsual?

The sample mean will not be considered unsual because it has a probability that is greater than 5%