4.3: Using the Normal Distribution
 Page ID
 79019
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left#1\right}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The shaded area in the following graph indicates the area to the right of \(x\). This area is represented by the probability \(P(X > x)\). Normal tables  see Appendix A in Chapter 11.1  provide the probability above a specific value such as \(x_1\). This is the shaded part of the graph below.
Because the normal distribution is symmetrical, if \(x_1\) were the same distance to the left of the mean, the probability in the left tail (below that value), would be the same as the shaded area in the right tail as shown in the figure. In general, bear in mind that because of the symmetry of this distribution, onehalf of the probability is to the right of the mean and onehalf is to the left of the mean.
Calculations of Probabilities
Let's discuss how to find the probability of a specified region in a standard normal distribution. The shaded region in the figure below shows that the area between \(x_1\) and \(x_2\) is the probability as stated in the formula: \(P (X_1 \leq X \leq X_2)\). In this case, suppose we have \(\mu\) = 5 and \(\sigma\) = 2. Suppose that \(x_1\) = 7 and \(x_2\) = 9. We want to find the probability of a score falling between 7 and 9 on this distribution.
The solution is to convert the distribution we have with its mean and standard deviation to the Standard Normal Distribution. The Standard Normal has a random variable called \(Z\). To compute probabilities, areas, for any normal distribution, we need only to convert the particular normal distribution to the standard normal distribution and look up the answer in the tables. As review, here again is the standardizing formula:
\[z=\frac{x\mu}{\sigma}\nonumber\]
where \(z\) is the value on the standard normal distribution, \(x\) is the value from a normal distribution one wishes to convert to the standard normal, \(\mu\) and \(\sigma\) are, respectively, the mean and standard deviation of that population. Note that the equation uses \(\mu\) and \(\sigma\) which denotes population parameters. This is still dealing with probability so we always are dealing with the population, with known parameter values and a known distribution. It is also important to note that because the normal distribution is symmetrical it does not matter if the zscore is positive or negative when calculating a probability. One standard deviation to the left (negative zscore) covers the same area as one standard deviation to the right (positive zscore). This fact is why the Standard Normal tables do not provide areas for the left side of the distribution. Because of this symmetry, the zscore formula can also be written as:
\[Z=\frac{x\mu}{\sigma}\nonumber\]
where the vertical lines in the equation means the absolute value of the number.
Using this formula, we can determine that \(x_1\) = 7 converts to a zscore of +1, and \(x_2\) = 9 converts to a zscore of 2.
Then, using the \(z\) table, to find the probability of \(z\) = 1, go to the \(z\) column, reading down to 1.0 and then read at column 0. That number, \(0.1587\) is the probability of a score falling at or above \(z\) = 1. We repeat this process  using the \(z\) table  to find the probability of a score falling at or above \(z\) = 2, which is \(0.0228\). To obtain the shaded area that we wish to know about (as shown in Figure 4.3.2 above), we will need to subtract \(0.1587  0.0228\) to obtain our final probability, \(.1359\).
To compute probabilities, areas, for any normal distribution, we need only to convert the particular normal distribution to the standard normal distribution using the zformula and look up the answer in the tables.
What the standardizing formula is really doing is computing the number of standard deviations \(x\) is from the mean of its own distribution. The standardizing formula and the concept of counting standard deviations from the mean is the secret of all that we will do in this statistics class. The reason this is true is that all of statistics boils down to variation, and the counting of standard deviations is a measure of variation.
This formula, in many disguises, will reappear over and over throughout this course.
Example \(\PageIndex{1}\)
The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five.
a. Find the probability that a randomly selected student scored more than 65 on the exam.
b. Find the probability that a randomly selected student scored less than 85.

a. Let \(x\) = a score on the final exam. \(X \sim N(63, 5)\), where \(\mu = 63\) and \(\sigma = 5\).
Draw a graph.
Then, find \(P(x > 65)\).
\(P(x > 65) = 0.3446\)
\[z_{1}=\frac{x_{1}\mu}{\sigma}=\frac{6563}{5}=0.4\nonumber\]
Looking up this value of z in our ztable, \(P\left(x \geq x_{1}\right)=P\left(z \geq z_{1}\right)=0.3446\)
The probability that any student selected at random scores more than 65 is 0.3446.
Answer
 Answer

b.
\(z=\frac{x\mu}{\sigma}=\frac{8563}{5}=4.4\). The closest value in our ztable is 4.5, which has a probability of .00000340.
Therefore, the probability that one student scores less than 85 is approximately one or 100% (i.e., 1  the probability of scoring above that value).
Exercise \(\PageIndex{1}\)
The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three.
Find the probability that a randomly selected golfer scored less than 65.
Example \(\PageIndex{2}\)
A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times for entertainment are normally distributed and the standard deviation for the times is half an hour.
a. Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day.

a. Let \(x\) = the amount of time (in hours) a household personal computer is used for entertainment. \(X \sim N(2, 0.5)\) where \(\mu= 2\) and \(\sigma = 0.5\).
Find \(P(1.8 < x < 2.75)\).
The probability for which you are looking is the area between \(x = 1.8\) and \(x = 2.75\). \(P(1.8 < x < 2.75) = 0.5886\)
\(P(1.8 \leq x \leq 2.75) = P(z_1 \leq z \leq z_2)\)
The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886.
Answer
b. Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment.

b. To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, first find the 25^{th} percentile, \(k\), where \(P(x < k) = 0.25\).
\(f(Z)=0.50.25=0.25, \text { therefore } z \approx0.675(\text { or just } 0.68 \text { using the table) } z=\frac{x\mu}{\sigma}=\frac{x2}{0.5}=0.675 , \text {therefore } x=0.675 * 0.5+2=1.66\)
The maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours.
Answer
Exercise \(\PageIndex{2}\)
The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a golfer scored between 66 and 70.
Example \(\PageIndex{3}\)
In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively.
a. Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old.

a. 0.8186
Answer
b. Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old.

b. 0.8413
Answer
Example \(\PageIndex{4}\)
A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm.
a. Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Sketch the graph.

\[z_{1}=\frac{65.85}{.24}=.625\nonumber\]
\(P(x \geq 6) = P(z \geq 0.625) = 0.2670\)
Answer
b. The middle 20% of mandarin oranges from this farm have diameters between ______ and ______.

\(f(z)=\frac{0.20}{2}=0.10, \text { therefore } z \approx \pm 0.25\)
\(z=\frac{x\mu}{\sigma}=\frac{x5.85}{0.24}=\pm 0.25 \rightarrow \pm 0.25 \cdot 0.24+5.85=(5.79,5.91)\)
Answer