Question

Question 9 (1 point)

The histograms and summary statistics summarize the data for the number of hits in the season by baseball players

in two leagues. Each data set contains one outlier. What are the values of the two outliers? Explain how each value

is determined to be an outlier.

Some summary statistics for the number of hits by players in each league.

mean median standard deviation minimum Q1 Q3

maximum

league A 151.12 148 26.83

29

136 167 207

league B 163.25 157 24.93

136 145 178 256



Plsss helppp

Answer 1

Explanation:

League A League B

151.12 163.25

148 157

26.83 24.93

29 136

136 145

167 178

207 256

League A in ascending order :

26.83 , 29 , 136, 148 , 151.12 , 167,207

`Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = (26.83+29 +136+ 148+ 151.12+ 167+207)/(7)\\\\Mean =123.564`

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=148

Standard deviation=
`\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}`=
`=\sqrt{((26.83-123.564)^2+(29-123.564)^2+.......+(207-123.564)^2)/(7)}=63.98`

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 26.83 , 29 , 136, 148

n = 4

Q1=82.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 148 , 151.12 , 167,207

n = 4

Q3=159.06

IQR = Q3-Q1=159.06-82.5=76.56

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

`(82.5-1.5* 76.56,159.06+1.5* 76.56)`

(-32.34,273.9)

So, There is no outlier

Maximum = 207

2)

League B in ascending order :

24.93,136,145,157,163.25,178,256

`Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = (24.93+136+145+157+163.25+178+256)/(7)\\\\Mean =151.45`

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=157

Standard deviation=
`\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}`=
`=\sqrt{((24.93-151.45)^2+(136-151.45)^2+.......+(256-151.45)^2)/(7)}=68.42`

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 24.93,136,145,157

n = 4

`Median = \frac{(n)/(2) \text{th term}+((n)/(2)+1) \text{th term}}{2}\\Median = \frac{(4)/(2) \text{th term}+((4)/(2)+1) \text{th term}}{2}\\Median = \frac{2 \text{th term}+3 \text{th term}}{2}\\Median = (136+145)/(2)=140.5`

Q1=140.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 157,163.25,178,256

n = 4

Q3=170.625

IQR = Q3-Q1=170.625-140.5=30.125

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

`(140.5-1.5* 30.125,170.625+1.5* 30.125)`

(95.3125,215.8125)

24.93 and 256 are outliers

Maximum = 256