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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 meanx¯.

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 40thpercentile.

k. Find the 90thpercentile.

l. Construct a line graph of the data.

m. Construct a stemplot of the data.

Short Answer

Expert verified

(a) Sample mean 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) 40thpercentile = 4.

(k) 90thpercentile = 5.

(l) Line graph.

(m) The stemplot,

Step by step solution

01

Mean :

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

02

; (a) Explanation :

Sample mean,

x¯=∑fx∑fx¯=15140=3.775

Hence, sample meanx¯=3.775.

03

(b) Explanation : 

s2=∑fx-x¯2n-1

n = number of data points,

x = each value of data,

x¯= mean of x,

s= standard deviation.

Substituting the values,

s2=∑fx-x¯2n-1=64.97539=1.6660

Sample standard deviation s=1.6660=1.290.

Hence, the standard deviation is1.290.

04

(c) Explanation : 

The graph is not symmetric.

05

(d) Explanation : 

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

06

(e) Explanation : 

First quartile position = n+1×25100=n+1×14=424=10.5.

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

10.5thposition value = 3+32=3.

Hence, the first quartile is3.

07

(f) Explanation : 

Median position = n+1×50100=n+1×12=422=21

21stposition value = 4.

Hence, median =4.

08

(g) Explanation : 

Third quartile position = n+1×75100=n+1×3×424=424=31.5

31.5is between 31stand 32ndposition which has value 5.

role="math" localid="1647961387124" 31.5thposition value = 5+52=5.

Hence, third quartile =5.

09

(h) Explanation : 

Min =1,

Max =7,

First quartile =3,

Median =4,

Third quartile = 5.

These values help us to draw the box plot.

10

(i) Explanation : 

When we add relative frequencies from 5, we get

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

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

11

(j) Explanation : 

40thpercentile - notice the 0.40in the cumulative relative frequency column.

The 40thpercentile is between 3and 4. But X values are not between 3and 4.

Hence,40thpercentile =4.

12

(k) Explanation : 

90thpercentile - notice the 0.90in cumulative relative frequency column.

The 90thpercentile is between the 4and 5. But X values are not between 4and 5.

From 67.5thpercentile to 97.5thpercentile, all X = 5.

Hence,90thpercentile =5.

13

(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.

14

(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.

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