Question

For the standard normal curve, find the z-score(s)

a. that has area $0.30$to its left.

b. that has area $0.10$to its right.

c.$z_{0.025,}{z}_{0.05},z_{0.01}$, and $z_{0005}$.

d. that divide the area under the curve into a middle $0.99$area and two outside $0.005$areas.

Short answer

(a) $Z$score for $0.30$area to the left $=-0.524$

(b) $Z$score for the $0.10$area to its right. $=1.282$

(c)

$\begin{array}{|llll|}\hline \text{Area to the right}& \text{Area to the left}& \text{Percentile}& \text{Z score}\\ Z_{0.025}& 1-0.025=0.975& 97.5& 1.96\\ Z_{0.05}& 1-0.05=0.95& 95& 1.645\\ Z_{0.01}& 1-0.01=0.99& 99& 2.326\\ Z_{0.005}& 1-0.005=0.995& 99.5& 2.576\\ \hline\end{array}$

(d)

$\begin{array}{|llll|}\hline Z_{0.005}& 1-0.005=0.995& 99.5& 2.576\\ Z_{0.995}& 1-0.995=0.005& 0.5& -2.576\\ \hline\end{array}$

Step-by-step solution

Part (a)Step 1: Given information

Given in the question that, we need to find the z-score(s) for the standard normal curve that has area 0.30 to its left.

Part(a) Step 2: Explanation

The data is normally distributed, and the area to its left is $0.30$. The area $0.30$to its left denotes the${30}^{\text{th}}$percentile.

The inverse normal distribution function can be used to compute the percentile's $z$-score:

- $($percentile $/100)=$z score InvNorm

We acquire the result for $z$score for that instance by filling in the percentile values in the function.

- InvNorm $(0.3)=-0.524$

- Z score for $0.30$area to the left $=-0.524$

Part(b) Step 1: Given information

Given in the question that, we need to find the z-score(s) for the standard normal curve that has area $0.10$to its left.

Part (b) Step 2: Explanation

The data is regularly distributed, and the region to the right of it is $0.10$. The area $0.10$to its right represents the ${90}^{\text{th}}$percentile.

The inverse normal distribution function can be used to compute the percentile's $z$-score: -

InvNorm(percentile/100) $=$z score

We acquire the result for $z$score for that instance by filling in the percentile values in the function.

- InvNorm$(0.9)=$$1.282$- $Z$score for the $0.10$right-hand area $=1.282$

Part(c) Step 1: Given information

For the standard normal curve,we need to find We need to find the z-score(s)

$z_{0.025,}{z}_{0.05},z_{0.01}$, and $z_{0005}$.

Part(c) Step 2: Explanation

The information is generally dispersed.

$\begin{array}{|lll|}\hline \text{Area to the right}& \text{Area to the left}& \text{Percentile}\\ {\mathrm{Z}}_{0.025}& 1-0.025=0.975& 97.5\\ {\mathrm{Z}}_{0.05}& 1-0.05=0.95& 95\\ {\mathrm{Z}}_{0.01}& 1-0.01=0.99& 99\\ {\mathrm{Z}}_{0.005}& 1-0.005=0.995& 99.5\\ \hline\end{array}$

The inverse normal distribution function can be used to calculate the percentile's z-score:

InvNorm(percentile$/100)$ = z score

We retrieve the result for $z$score for that instance by plugging in the percentile values into the function.

$\begin{array}{|llll|}\hline \text{Area to the right}& \text{Area to the left}& \text{Percentile}& \text{Z score}\\ Z_{0.025}& 1-0.025=0.975& 97.5& 1.96\\ Z_{0.05}& 1-0.05=0.95& 95& 1.645\\ Z_{0.01}& 1-0.01=0.99& 99& 2.326\\ Z_{0.005}& 1-0.005=0.995& 99.5& 2.576\\ \hline\end{array}$

Part(d) Step 1: Given information

For the standard normal curve, We need to find the z-score(s) that divide the area under the curve into a middle $0.99$area and two outside $0.005$areas.

Part (d) Step 2: Explanation

Assume that the data is regularly distributed.

Two curves with area under curves of $0.99$in the middle and $0.005$on the sides.

The inverse normal distribution function can be used to calculate the percentile's z-score:

InvNorm $($percentile $/100)=$ z score

We acquire the result for $z$score for that instance by filling in the percentile values in the function.

$\begin{array}{|llll|}\hline Z_{0.005}& 1-0.005=0.995& 99.5& 2.576\\ Z_{0.995}& 1-0.995=0.005& 0.5& -2.576\\ \hline\end{array}$