2) Compute: \( \mathrm{p}(0.8<z<2.19) \). You must sketch, shade, and label a bell curve for credit. 3) Find the missing z-number: \( \mathrm{p}(\mathrm{z}<\mathrm{c})=0.7 \). You must sketch, shade, and label a bell curve for credit.

A random sample of 69 produced the summary statistics \( \bar{x}=0.326 \) and \( s^{2}=0.064 \). Complete parts a and \( \mathbf{b} \). b. Test the null hypothesis that \( \mu=0.37 \) against the alternative hypothesis that \( \mu \neq 0.37 \), using \( \alpha=0.10 \). What is the value of the test statistic? F=-1.44 (Round to two decimal places as needed.) Find the p-value. (Round to three decimal places as needed.)

A variable \( x \) is normally distributed with mean 22 and standard deviation 3. Round your answers to the nearest hundredth as needed. a) Determine the \( z \)-score for \( x=27 \). b) Determine the \( z \)-score for \( x=20 \). c) What value of \( x \) has a \( z \)-score of 2.33 ? d) What value of \( x \) has a \( z \)-score of -0.3 ? \( x= \) e) What value of \( x \) has a \( z \)-score of 0 ?

a. What is the difference between the number who prefer the cola to the lemon flavour? b. How many Grade 4D learners were questioned altogether?

\( 1 \leftarrow \begin{array}{l}\text { Researchers conducted a study on using a chief executive officer's facial structure to predict a firm's financial } \\ \text { performance. The facial width-to-height ratio (WHR) for each in a sample of } 52 \text { CEOs at publicly traded firms was } \\ \text { determined. The sample resulted in } \bar{x}=1.55 \text { and } s=0.36 \text {. An analyst wants to predict the financial performance of a } \\ \text { firm based on the value of the true mean facial WHR of CEOs. The analyst wants to use the value of } \mu=2.1 \text {. Do yo } \\ \text { recommend he use this value? Conduct a test of hypothesis for } \mu \text { to help you answer the question. Specify all the } \\ \text { elements of the test, including } H_{0}, H_{a} \text {, test statistic, } p \text {-value, and your conclusion. Test at } \alpha=0.01 \text {. } \\ \text { What is the value of the test statistic? } \\ z=-11.02 \text { (Round to two decimal places as needed.) } \\ \text { Find the p-value. } \\ \text { p-value }=\square \text { (Round to three decimal places as needed.) }\end{array} \).

In a normal distribution, a data value located 1.2 standard deviations below the mean has Standard Score: \( \mathrm{z}=\square \)