Normal Distribution

According to the website, the mean is $183$ and the standard deviation is $5.$ Therefore, $\mu =183$ and $\sigma =5.$ Consequently, the middle $68\%$ of the distribution represents the range of heights from $178$ to $188$ centimeters. Now, notice that $173$ centimeters is $2$ standard deviations to the left of the mean. 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 $\mu +2\sigma .$ 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.

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

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. 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 $0.12.$

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 $P(\text{-}0.4<z<0.8).$ 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. The probability that Kevin's commute will take between $16$ and $19$ minutes next Monday is about $0.44.$

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 $1.$ 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 $P(z>0.8).$ The probability that Kevin will be late for work on that day is about $0.22.$

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 $z=0.6.$

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