# How to Use the Z-table to Compute Probabilities of Non-Standard Normal Distributions

## Problem 758

Let $X\sim \mathcal{N}(\mu, \sigma)$ be a normal random variable with parameter $\mu=6$ and $\sigma^2=4$. Find the following probabilities using the Z-table below.

(a) Find $P(X \lt 7)$.

(b) Find $P(X \lt 3)$.

(c) Find $P(4.5 \lt X \lt 8.5)$.

## Solution.

To make use of the Z-table, we need to relate current problems to a standard normal distribution $\mathcal{N}(0, 1)$ with mean $0$ and deviation $1$. To do this, as $X$ is a normal random variable with mean $\mu = 6$ and standard deviation $\sigma = 2$, the new random variable $Z$ defined by
$Z = \frac{X – 6}{2}$ is a standard normal random variable.

### Solution of (a)

The inequality $X < 7$ is equivalent to the inequality $Z = \frac{X - 6}{2} < \frac{7-6}{2} = 0.5.$ Thus, the required probability is \begin{align*} P(X < 7) = P(Z < 0.5)= \Phi(0.5). \end{align*} Here, $\Phi(x)$ is the cumulative distribution function of a standard normal random variable $Z$. The value of $\Phi(0.5)$ can be found from the Z-table. Looking at row 0.5 and column 0.00, we see that $\Phi(0.5) \approx 0.6915$. Hence, the answer is
$P(X < 7 ) \approx 0.6915.$

### Solution of (b)

As we did in Part (a), we transform the inequality and get
\begin{align*}
P(X \lt 3) &= P\left(\frac{X-6}{2} \lt \frac{3-6}{2}\right)\6pt] &= P(Z \lt -1.5)\\ &= \Phi(-1.5). \end{align*} Now, note that the Z-table gives the values of \Phi(x) for only non-negative x. Thus, to compute \Phi(-1.5), we use the symmetry of the graph of a standard normal distribution. As \Phi(-1.5) is the area under the bell curve from -\infty to -1.5, this is equal to the area under the bell curve from 1.5 to \infty by symmetry, which is the same as 1-\Phi(1.5). See the figure below. In the figure, the left orange region is \Phi(-1.5), which is equal to the right orange region. Since the total area under the curve \Phi(x) is 1, the area of the right orange region is 1 – \Phi(1.5). Thus \begin{align*} \Phi(-1.5) &= 1 – \Phi(1.5)\\ &\approx 1 – 0.9332 && \text{ from the Z-table}\\ &= 0.0668. \end{align*} Thus, we obtain the probability \[P(X < 3) \approx 0.0668.

### Solution of (c)

Again, by normalizing, we obtain
\begin{align*}
P(4.5 < X < 8.5) &= P\left( \frac{4.5 - 6}{2} < X < \frac{X - 6}{2} < \frac{8.5 - 6}{2} \right)\\ &=P(-0.75 < Z < 1.25)\\ &= \Phi(1.25) - \Phi(-0.75). \end{align*} Now, to find the value of $\Phi(-0.75)$ from the Z-table, we use the symmetry of the bell curve as in part (b) and we see $\Phi(-0.75) = 1 – \Phi(0.75)$.
Thus,
\begin{align*}
P(4.5 < X < 8.5) &= \Phi(1.25) - \Phi(-0.75)\\ &= \Phi(1.25) - (1-\Phi(0.75))\\ &= \Phi(1.25) + \Phi(0.75) - 1\\ &\approx 0.8944 + 0.7734 - 1\\ &=0.6678. \end{align*} Note that we found values $\Phi(1.25)\approx 0.8944$ and $\phi(0.75)\approx 0.7734$ from the Z-table.
In conclusion, we have
$P(4.5 < X < 8.5) \approx 0.6678.$

## Z-table

The Z-table below gives numerical values for the cummulative distribution function $\Phi(x)$ of the standard normal random variable $Z$.

(You may need a large display to see the whole table.)

\begin{array}{rrrrrrrrrrr}
\hline
& 0.00 & 0.01 & 0.02 & 0.03 & 0.04 & 0.05 & 0.06 & 0.07 & 0.08 & 0.09 \\
\hline
0.0 & 0.5000 & 0.5040 & 0.5080 & 0.5120 & 0.5160 & 0.5199 & 0.5239 & 0.5279 & 0.5319 & 0.5359 \\
0.1 & 0.5398 & 0.5438 & 0.5478 & 0.5517 & 0.5557 & 0.5596 & 0.5636 & 0.5675 & 0.5714 & 0.5753 \\
0.2 & 0.5793 & 0.5832 & 0.5871 & 0.5910 & 0.5948 & 0.5987 & 0.6026 & 0.6064 & 0.6103 & 0.6141 \\
0.3 & 0.6179 & 0.6217 & 0.6255 & 0.6293 & 0.6331 & 0.6368 & 0.6406 & 0.6443 & 0.6480 & 0.6517 \\
0.4 & 0.6554 & 0.6591 & 0.6628 & 0.6664 & 0.6700 & 0.6736 & 0.6772 & 0.6808 & 0.6844 & 0.6879 \\
0.5 & 0.6915 & 0.6950 & 0.6985 & 0.7019 & 0.7054 & 0.7088 & 0.7123 & 0.7157 & 0.7190 & 0.7224 \\
0.6 & 0.7257 & 0.7291 & 0.7324 & 0.7357 & 0.7389 & 0.7422 & 0.7454 & 0.7486 & 0.7517 & 0.7549 \\
0.7 & 0.7580 & 0.7611 & 0.7642 & 0.7673 & 0.7704 & 0.7734 & 0.7764 & 0.7794 & 0.7823 & 0.7852 \\
0.8 & 0.7881 & 0.7910 & 0.7939 & 0.7967 & 0.7995 & 0.8023 & 0.8051 & 0.8078 & 0.8106 & 0.8133 \\
0.9 & 0.8159 & 0.8186 & 0.8212 & 0.8238 & 0.8264 & 0.8289 & 0.8315 & 0.8340 & 0.8365 & 0.8389 \\
1.0 & 0.8413 & 0.8438 & 0.8461 & 0.8485 & 0.8508 & 0.8531 & 0.8554 & 0.8577 & 0.8599 & 0.8621 \\
1.1 & 0.8643 & 0.8665 & 0.8686 & 0.8708 & 0.8729 & 0.8749 & 0.8770 & 0.8790 & 0.8810 & 0.8830 \\
1.2 & 0.8849 & 0.8869 & 0.8888 & 0.8907 & 0.8925 & 0.8944 & 0.8962 & 0.8980 & 0.8997 & 0.9015 \\
1.3 & 0.9032 & 0.9049 & 0.9066 & 0.9082 & 0.9099 & 0.9115 & 0.9131 & 0.9147 & 0.9162 & 0.9177 \\
1.4 & 0.9192 & 0.9207 & 0.9222 & 0.9236 & 0.9251 & 0.9265 & 0.9279 & 0.9292 & 0.9306 & 0.9319 \\
1.5 & 0.9332 & 0.9345 & 0.9357 & 0.9370 & 0.9382 & 0.9394 & 0.9406 & 0.9418 & 0.9429 & 0.9441 \\
1.6 & 0.9452 & 0.9463 & 0.9474 & 0.9484 & 0.9495 & 0.9505 & 0.9515 & 0.9525 & 0.9535 & 0.9545 \\
1.7 & 0.9554 & 0.9564 & 0.9573 & 0.9582 & 0.9591 & 0.9599 & 0.9608 & 0.9616 & 0.9625 & 0.9633 \\
1.8 & 0.9641 & 0.9649 & 0.9656 & 0.9664 & 0.9671 & 0.9678 & 0.9686 & 0.9693 & 0.9699 & 0.9706 \\
1.9 & 0.9713 & 0.9719 & 0.9726 & 0.9732 & 0.9738 & 0.9744 & 0.9750 & 0.9756 & 0.9761 & 0.9767 \\
2.0 & 0.9772 & 0.9778 & 0.9783 & 0.9788 & 0.9793 & 0.9798 & 0.9803 & 0.9808 & 0.9812 & 0.9817 \\
2.1 & 0.9821 & 0.9826 & 0.9830 & 0.9834 & 0.9838 & 0.9842 & 0.9846 & 0.9850 & 0.9854 & 0.9857 \\
2.2 & 0.9861 & 0.9864 & 0.9868 & 0.9871 & 0.9875 & 0.9878 & 0.9881 & 0.9884 & 0.9887 & 0.9890 \\
2.3 & 0.9893 & 0.9896 & 0.9898 & 0.9901 & 0.9904 & 0.9906 & 0.9909 & 0.9911 & 0.9913 & 0.9916 \\
2.4 & 0.9918 & 0.9920 & 0.9922 & 0.9925 & 0.9927 & 0.9929 & 0.9931 & 0.9932 & 0.9934 & 0.9936 \\
2.5 & 0.9938 & 0.9940 & 0.9941 & 0.9943 & 0.9945 & 0.9946 & 0.9948 & 0.9949 & 0.9951 & 0.9952 \\
2.6 & 0.9953 & 0.9955 & 0.9956 & 0.9957 & 0.9959 & 0.9960 & 0.9961 & 0.9962 & 0.9963 & 0.9964 \\
2.7 & 0.9965 & 0.9966 & 0.9967 & 0.9968 & 0.9969 & 0.9970 & 0.9971 & 0.9972 & 0.9973 & 0.9974 \\
2.8 & 0.9974 & 0.9975 & 0.9976 & 0.9977 & 0.9977 & 0.9978 & 0.9979 & 0.9979 & 0.9980 & 0.9981 \\
2.9 & 0.9981 & 0.9982 & 0.9982 & 0.9983 & 0.9984 & 0.9984 & 0.9985 & 0.9985 & 0.9986 & 0.9986 \\
3.0 & 0.9987 & 0.9987 & 0.9987 & 0.9988 & 0.9988 & 0.9989 & 0.9989 & 0.9989 & 0.9990 & 0.9990 \\
\hline
\end{array}

Let $X$ be an exponential random variable with parameter $\lambda$. (a) For any positive integer $n$, prove that \[E[X^n] =...