Question

Forty randomly selected students were asked the number of pairs of sneakers they owned. Let $X$= the number of pairs of sneakers owned. The results are as follows:

a. Find the sample mean$\overline{x}$.

b. Find the sample standard deviation, s

c. Construct a histogram of the data.

d. Complete the columns of the chart.

e. Find the first quartile.

f. Find the median.

g. Find the third quartile.

h. Construct a box plot of the data.

i. What percent of the students owned at least five pairs?

j. Find the ${40}^{th}$percentile.

k. Find the ${90}^{th}$percentile.

l. Construct a line graph of the data.

m. Construct a stemplot of the data.

Short answer

(a) Sample mean $\overline{x}=3.775$.

(b) Sample standard deviation $s=1.290$.

(c) Histogram of data.

(d) The columns,

(e) First quartile = $3$.

(f) Median = $4$.

(g) Third quartile = $5$.

(h) Box plot.

(i) Students owned at least five pairs = $32.5\%$.

(j) ${40}^{th}$percentile = $4$.

(k) ${90}^{th}$percentile = $5$.

(l) Line graph.

(m) The stemplot,

Step-by-step solution

Mean :

The total of all observation or data point values divided by the number of observations.

; (a) Explanation :

Sample mean,

$$\begin{gathered} \overline{x}=\frac{\sum _{}fx}{\sum _{}f} \\ \overline{x}=\frac{151}{40}=3.775 \end{gathered}$$

Hence, sample mean$\overline{x}=3.775$.

(b) Explanation : 

$s^2=\frac{\sum _{}f{\left(x-\overline{x}\right)}^2}{n-1}$

$n$ = number of data points,

$x$ = each value of data,

$\overline{x}$= mean of x,

$s$= standard deviation.

Substituting the values,

$s^2=\frac{\sum _{}f{\left(x-\overline{x}\right)}^2}{n-1}=\frac{64.975}{39}=1.6660$

Sample standard deviation $s=\sqrt{1.6660}=1.290$.

Hence, the standard deviation is$1.290$.

(c) Explanation : 

The graph is not symmetric.

(d) Explanation : 

Divide each frequencies by the total number of observations in the sample to find relative frequencies. In this case,$40$.

(e) Explanation : 

First quartile position = $\left(n+1\right)\times \frac{25}{100}=\left(n+1\right)\times \frac{1}{4}=\frac{42}{4}=10.5$.

Value $10.5$is between tenth and eleventh position that had value $3$.

$10.5^{th}$position value = $\frac{3+3}{2}=3$.

Hence, the first quartile is$3$.

(f) Explanation : 

Median position = $\left(n+1\right)\times \frac{50}{100}=\left(n+1\right)\times \frac{1}{2}=\frac{42}{2}=21$

${21}^{st}$position value = $4$.

Hence, median =$4$.

(g) Explanation : 

Third quartile position = $\left(n+1\right)\times \frac{75}{100}=\left(n+1\right)\times \frac{3\times 42}{4}=\frac{42}{4}=31.5$

$31.5$is between ${31}^{st}$and ${32}^{nd}$position which has value $5$.

role="math" localid="1647961387124" $31.5^{th}$position value = $\frac{5+5}{2}=5$.

Hence, third quartile =$5$.

(h) Explanation : 

Min =$1$,

Max =$7$,

First quartile =$3$,

Median =$4$,

Third quartile = $5$.

These values help us to draw the box plot.

(i) Explanation : 

When we add relative frequencies from $5$, we get

$0.3+0+0.025=0.325$of frequency. That is $32.5\%$.

Hence, $32.5\%$of the students owned at least five pairs.

(j) Explanation : 

${40}^{th}$percentile - notice the $0.40$in the cumulative relative frequency column.

The ${40}^{th}$percentile is between $3$and $4$. But X values are not between $3$and $4$.

Hence,${40}^{th}$percentile =$4$.

(k) Explanation : 

${90}^{th}$percentile - notice the $0.90$in cumulative relative frequency column.

The ${90}^{th}$percentile is between the $4$and $5$. But X values are not between $4$and $5$.

From $67.5^{th}$percentile to $97.5^{th}$percentile, all X = $5$.

Hence,${90}^{th}$percentile =$5$.

(l) Explanation : 

Line graph is another type of graph useful for specific data values.

The x-axis contains data values and the y-axis contains frequency points.

The frequency points are connected through line segments.

(m) Explanation : 

The stem plot is a faster way to graph data and provide an exact picture of data. To create the plot, divide each observation of data into stem and leaf. And leaf has a final significant digit.