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In his spare time, Kevin works with the Less Chat, More Talk campaign to encourage people to share with their loved ones in person instead of through screens. He wants to give away T-shirts with a cool logo outside a shopping mall to help spread this message.

Kevin is in charge of preparing the men's T-shirts, but he does not know how many of each size he should order. To figure it out, he searched the City Hall website and he found that the heights of the men in the city are normally distributed with a mean of $183$ centimeters and a standard deviation of $5$ centimeters. Along with this information, there was also a graph.

Solution

a According to the Empirical Rule, the middle $68\%$ of the data in a normal distribution falls in the range that starts one standard deviation to the left of the mean and ends one standard deviation to the right of the mean.

$$\begin{array}{c}\text{Middle }68\%\\ \mu -\sigma <X<\mu +\sigma \end{array}$$

According to the website, the mean is $183$ and the standard deviation is $5.$ Therefore, $\mu =183$ and $\sigma =5.$

$$\begin{array}{c}\text{Middle }68\%\\ 183-5<X<183+5\\ \Updownarrow \\ 178<X<188\end{array}$$

Consequently, the middle $68\%$ of the distribution represents the range of heights from $178$ to $188$ centimeters.

b Start by highlighting the corresponding interval in the graph.

Now, notice that $173$ centimeters is $2$ standard deviations to the left of the mean.

$$\begin{array}{rl}\mu -2\sigma & =183-2(5)\\ & \Updownarrow \\ \mu -2\sigma & =173\end{array}$$

According to the Empirical Rule, $95\%$ of the data fall between $\mu -2\sigma$ and $\mu +2\sigma .$ It is known that the value of $\mu -2\sigma$ is $173.$ Calculate the value of $\mu +2\sigma .$

$$\begin{array}{rl}\mu +2\sigma & =183+2(5)\\ & \Updownarrow \\ \mu +2\sigma & =193\end{array}$$

Therefore, $95\%$ of the data fall between $173$ and $193.$ This implies that $5\%$ of the data fall outside this range.

Due to the symmetry of the normal curve, $2.5\%$ of the data fall to the left of $173$ and $2.5\%$ of the data fall to the right of $193.$ Consequently, $2.5\%$ of the men surveyed are shorter than $173$ centimeters.

c The graph below shows the percentages represented by each interval according to the Empirical Rule.

According to the graph, $13.5\%$ of the surveyed men are between $188$ and $193$ centimeters tall.

To find the number of men that belong to this range, multiply the corresponding percentage by the total number of men that participated in the survey.

$13.5\%(3000)$

Multiply

PercentToFrac

$a\%=\frac{a}{100}$

$\frac{13.5}{100}(3000)$

MoveRightFacToNum

$\frac{a}{c}\cdot b=\frac{a\cdot b}{c}$

$\frac{40500}{100}$

CalcQuot

Calculate quotient

$405$

Therefore, of the $3000$ men surveyed, about $405$ are between $188$ and $193$ centimeters tall.

Kevin has become a stats fan. He has recorded the time it takes him to commute to his internship over the past few days. He observes that the times are normally distributed with a mean of $17$ minutes and a standard deviation of $2.5$ minutes.

Find the following probabilities and write them in decimal form rounded to two decimal places.

Solution

a Start by drawing the normal distribution curve. According to Kevin, the mean time it takes him to get to work is $17$ minutes, so this value should be located in the middle of the axis. The standard deviation is $2.5$ minutes. The labels of the axis are found by adding and subtracting integer multiples of the standard deviation to and from the mean.

The probability that Kevin spends less than $14$ minutes getting to work tomorrow is represented by the area below the curve that is to the left of $14.$

Since $14$ is not a label on the axis, the Empirical Rule cannot be used. Therefore, to find the area, first convert $x=14$ into its corresponding $z\text{-}$score.

$z=\frac{x-\mu }{\sigma }$

SubstituteValues

Substitute values

$z=\frac{14-17}{2.5}$

Evaluate right-hand side

SubTerm

Subtract term

$z=\frac{\text{-}3}{2.5}$

MoveNegNumToFrac

Put minus sign in front of fraction

$z=\text{-}\frac{3}{2.5}$

CalcQuot

Calculate quotient

$z=\text{-}1.2$

Now, in the standard normal table, locate the row corresponding to $\text{-}1$ and the column corresponding to $.2.$

	$.0$	$.1$	$.2$	$.3$	$.4$	$.5$	$.6$	$.7$	$.8$	$.9$
$\text{-}3$	$.00135$	$.00097$	$.00069$	$.00048$	$.00034$	$.00023$	$.00016$	$.00011$	$.00007$	$.00005$
$\text{-}2$	$.02275$	$.01786$	$.01390$	$.01072$	$.00820$	$.00621$	$.00466$	$.00347$	$.00256$	$.00187$
$\text{-}1$	$.15866$	$.13567$	$.11507$	$.09680$	$.08076$	$.06681$	$.05480$	$.04457$	$.03593$	$.02872$
$\text{-}0$	$.50000$	$.46017$	$.42074$	$.38209$	$.34458$	$.30854$	$.27425$	$.24196$	$.21186$	$.18406$
$0$	$.50000$	$.53983$	$.57926$	$.61791$	$.65542$	$.69146$	$.72575$	$.75804$	$.78814$	$.81594$
$1$	$.84134$	$.86433$	$.88493$	$.90320$	$.91924$	$.93319$	$.94520$	$.95543$	$.96407$	$.97128$
$2$	$.97725$	$.98214$	$.98610$	$.98928$	$.99180$	$.99379$	$.99534$	$.99653$	$.99744$	$.99813$
$3$	$.99865$	$.99903$	$.99931$	$.99952$	$.99966$	$.99977$	$.99984$	$.99989$	$.99993$	$.99995$

According to the table, the probability that tomorrow Kevin will spend less than $14$ minutes traveling to work is about $0.12.$

b In the graph from Part A it can be seen that neither $16$ nor $19$ are labels on the axis.

Therefore, both values will need to be converted into their corresponding $z\text{-}$scores first. Recall that $\mu =17$ and $\sigma =2.5!$

$z=\frac{x-\mu }{\sigma }$
$x\text{-}$value	Substitute	Simplify
$16$	$z=\frac{16-17}{2.5}$	$z=\text{-}0.4$
$19$	$z=\frac{19-17}{2.5}$	$z=0.8$

The shaded area represents the probability that a randomly selected value is greater than $\text{-}0.4$ and smaller than $0.8.$ In other words, the shaded area represents $P(\text{-}0.4<z<0.8).$

$$\begin{array}{c}P(\text{-}0.4<z<0.8)\\ =\\ P(z<0.8)-P(z<\text{-}0.4)\end{array}$$

Each of these probabilities can be found using the standard normal table.

	$.0$	$.1$	$.2$	$.3$	$.4$	$.5$	$.6$	$.7$	$.8$	$.9$
$\text{-}3$	$.00135$	$.00097$	$.00069$	$.00048$	$.00034$	$.00023$	$.00016$	$.00011$	$.00007$	$.00005$
$\text{-}2$	$.02275$	$.01786$	$.01390$	$.01072$	$.00820$	$.00621$	$.00466$	$.00347$	$.00256$	$.00187$
$\text{-}1$	$.15866$	$.13567$	$.11507$	$.09680$	$.08076$	$.06681$	$.05480$	$.04457$	$.03593$	$.02872$
$\text{-}0$	$.50000$	$.46017$	$.42074$	$.38209$	$.34458$	$.30854$	$.27425$	$.24196$	$.21186$	$.18406$
$0$	$.50000$	$.53983$	$.57926$	$.61791$	$.65542$	$.69146$	$.72575$	$.75804$	$.78814$	$.81594$
$1$	$.84134$	$.86433$	$.88493$	$.90320$	$.91924$	$.93319$	$.94520$	$.95543$	$.96407$	$.97128$
$2$	$.97725$	$.98214$	$.98610$	$.98928$	$.99180$	$.99379$	$.99534$	$.99653$	$.99744$	$.99813$
$3$	$.99865$	$.99903$	$.99931$	$.99952$	$.99966$	$.99977$	$.99984$	$.99989$	$.99993$	$.99995$

Finally, to find the probability that Kevin will arrive within this time frame, calculate their difference and round the answer to two decimal places.

$P(\text{-}0.4<z<0.8)=P(z<0.8)-P(z<\text{-}0.4)$

SubstituteII

$P(z<0.8)=0.78814$, $P(z<\text{-}0.4)=0.34458$

$P(\text{-}0.4<z<0.8)=0.78814-0.34458$

SubTerm

Subtract term

$P(\text{-}0.4<z<0.8)=0.44356$

RoundDec

Round to $2$ decimal place(s)

$P(\text{-}0.4<z<0.8)\approx 0.44$

The probability that Kevin's commute will take between $16$ and $19$ minutes next Monday is about $0.44.$

c In order for Kevin to be on time, his commute cannot take more than $19$ minutes. In other words, he will be late for work if it takes more than $19$ minutes. This means that the probability that Kevin will be late for work that day is represented by the area below the normal curve that is to the right of $19.$

Since $19$ is not a label on the axis, the Empirical Rule cannot be used. Therefore, $z\text{-}$scores must be used to find the area. In Part B it was determined the $z\text{-}$score that corresponds to $19$ is $0.8.$

Probability of Kevin Being Late	Probability of Kevin Being on Time
$P(z>0.8)$	$P(z\le 0.8)$

Since the event of Kevin being late is the complement of the event of Kevin being on time, the sum of these probabilities is equal to $1.$

$$\begin{array}{c}P(z>0.8)+P(z\le 0.8)=1\end{array}$$

It was also determined in Part B that $P(z\le 0.8)$ is $0.78814.$ Substitute this value into the equation above and solve for $P(z>0.8).$

$P(z>0.8)+P(z\le 0.8)=1$

Substitute

$P(z\le 0.8)=0.78814$

$P(z>0.8)+0.78814=1$

SubEqn

$\text{LHS}-0.78814=\text{RHS}-0.78814$

$P(z>0.8)=0.22186$

RoundDec

Round to $2$ decimal place(s)

$P(z>0.8)\approx 0.22$

The probability that Kevin will be late for work on that day is about $0.22.$

The company Kevin is interning with plans to release a new smartphone. He goes with the research team to a stadium with a prototype to let different people use the phone in order to determine what features and design people like.

After comparing and contrasting size preference with the ages of the participants, Kevin realizes that the data is normally distributed. Additionally, he notices that the middle $46\%$ of participants prefer a larger phone.

Solution

a Let $z_1$ and $z_2$ be the lower and upper limits of the middle $46\%$ of the data. To find the corresponding values, start by finding the percentage of data outside the middle area. To do so, subtract $46\%$ from $100\%.$

Due to the symmetry of the normal curve, the area to the left of $z_1$ is equal to the area to the right of $z_2.$ Therefore, each portion corresponds to $54÷2=27\%$ of the data. For the moment, focus on the area to the left of $z_1.$

According to the last graph, the probability that a randomly chosen value is less than $z_1$ is $0.27.$ In other words, $P(z<z_1)=0.27.$ Now, look for the $z\text{-}$value that produces a probability of $0.27$ on a standard normal table.

	$.0$	$.1$	$.2$	$.3$	$.4$	$.5$	$.6$	$.7$	$.8$	$.9$
$\text{-}3$	$.00135$	$.00097$	$.00069$	$.00048$	$.00034$	$.00023$	$.00016$	$.00011$	$.00007$	$.00005$
$\text{-}2$	$.02275$	$.01786$	$.01390$	$.01072$	$.00820$	$.00621$	$.00466$	$.00347$	$.00256$	$.00187$
$\text{-}1$	$.15866$	$.13567$	$.11507$	$.09680$	$.08076$	$.06681$	$.05480$	$.04457$	$.03593$	$.02872$
$\text{-}0$	$.50000$	$.46017$	$.42074$	$.38209$	$.34458$	$.30854$	$.27425$	$.24196$	$.21186$	$.18406$
$0$	$.50000$	$.53983$	$.57926$	$.61791$	$.65542$	$.69146$	$.72575$	$.75804$	$.78814$	$.81594$
$1$	$.84134$	$.86433$	$.88493$	$.90320$	$.91924$	$.93319$	$.94520$	$.95543$	$.96407$	$.97128$
$2$	$.97725$	$.98214$	$.98610$	$.98928$	$.99180$	$.99379$	$.99534$	$.99653$	$.99744$	$.99813$
$3$	$.99865$	$.99903$	$.99931$	$.99952$	$.99966$	$.99977$	$.99984$	$.99989$	$.99993$	$.99995$

It is seen in the table that $z_1=\text{-}0.6.$ Again, due to symmetry, $z_2$ is the opposite of $z_1.$ Therefore, $z_2=0.6.$

Therefore, the limits of the middle $46\%$ of the data are $z=\text{-}0.6$ and $z=0.6.$

b In Part A it was determined that the limits of the middle $46\%$ of the data are $\text{-}0.6$ and $0.6.$

These $z\text{-}$scores can be converted back into their original age values in order to determine the range of ages that this area represents. To do so, start by rearranging the $z\text{-}$score formula to solve for $x.$

$z=\frac{x-\mu }{\sigma }$

Solve for $x$

MultEqn

$\text{LHS}\cdot \sigma =\text{RHS}\cdot \sigma$

$z\cdot \sigma =x-\mu$

AddEqn

$\text{LHS}+\mu =\text{RHS}+\mu$

$z\cdot \sigma +\mu =x$

RearrangeEqn

Rearrange equation

$x=\mu +z\cdot \sigma$

Kevin noted that the mean age is $\mu =19$ and the standard deviation is $\sigma =5.$ To find the first value of $x,$ substitute these values and $z=\text{-}0.6$ into the equation and simplify.

$x=\mu +z\cdot \sigma$

Substitute values and evaluate

SubstituteValues

Substitute values

$x=19+(\text{-}0.6)(5)$

MultNegPos

$(\text{-}a)b=\text{-}ab$

$x=19+(\text{-}3)$

AddNeg

$a+(\text{-}b)=a-b$

$x=19-3$

SubTerm

Subtract term

$x=16$

The age corresponding to $z=\text{-}0.6$ is $16.$ Now substitute $z=0.6$ to find the second age.

$x=\mu +z\cdot \sigma$

Substitute values and evaluate

SubstituteValues

Substitute values

$x=19+0.6(5)$

Multiply

Multiply

$x=19+3$

AddTerms

Add terms

$x=22$

The age corresponding to $z=0.6$ is $22.$ Therefore, the middle $46\%$ of the data corresponds to people between $16$ and $22$ years old.

Based on the Kevin's data, people between the ages of $16$ and $22$ prefer a larger phone.

$$\begin{array}{c}16<X<22\end{array}$$

Kevin's friend LaShay took the SAT and scored $640$ points on the math section. Kevin took the ACT and scored $28.32$ points in the math section.

Since these tests use different scales — the math section of the SAT scores $800$ points while the math section of the ACT scores $36$ points — they wonder who did better. They looked at the stats for each test to find out.

• LaShay's class scores are normally distributed with a mean of $523$ and a standard deviation of $90.$
• Kevin's class scores are also normally distributed with a mean of $21$ and a standard deviation of $6.1.$