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Central Tendency Worksheets
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1.
Paula and Irena traveled to Orlando this autumn. The number of miles they drove and the amount of gas utilized are given in the table. Calculate the average mileage (per gallon) of the car during the trip.
 Day 1 Day 2 Day 3 Day 4 Distance (in miles) 714 340 697 493 Gas utilized (in gallons) 42 20 41 29 a. 12 miles/gallon b. 27 miles/gallon c. 17 miles/gallon d. 22 miles/gallon

Solution:

Average mileage of the car during the trip = (Total number of miles traveled) / (Amount of gas utilized)

= (714+340+697+493)(42+20+41+29)
[Substitute the values.]

= 2244132 = 17 miles/gallon

So, the average mileage of the car during the trip was 17 miles/gallon

2.
The line graph shows the number of miles traveled by Paula each month. What is the month corresponding to the outlier in the line graph?  a. May b. June c. December d. November

Solution:

Outlier is the value which is much higher or lower than the rest of the data.

From the line graph, 700 is the outlier as it is much higher than the other data values.

So, the outlier is 700 miles, which is the distance traveled by Paula in the month of November.

3.
The loss percentages of a company for 6 months are given in the bar graph. What is the outlier in the data?  a. 6% b. 5% c. 4% d. 1%

Solution:

Outlier is the value that is much higher or lower than the other data.

In the graph, the percentage loss in the month of May is much lower than the other data values.

So, 1% is the outlier, which is the loss percentage in the month of May.

4.
What is the outlier in the box and whisker plot?  a. 92 b. 84 c. 90 d. No outlier

Solution:

In a box and whisker plot outlier is the data value that should be less than the [lower quartile - (1.5 x IQR)] or greater than the [upper quartile + (1.5 x IQR)].

= 100 - 90 = 10
IQR = upper quartile - lower quartile
[upper quartile = 100 and lower quartile = 90.]

Lower quartile - (1.5 x IQR) = 90 - (1.5 x 10) = 75
[Substitute and subtract.]

Upper quartile + (1.5 x IQR) = 100 + (1.5 x 10) = 115
[Substitute and simplify.]

In the box and whisker plot all the values are between 82 and 106. So, there is no value less than 75 or greater than 115.

So, in the data there is no outlier.

5.
The mean of the observations $\frac{1}{4}$, $\frac{1}{6}$, $\frac{3}{4}$, $\frac{1}{2}$, $\frac{5}{12}$, $\frac{2}{3}$ is: a. $\frac{33}{2}$ b. $\frac{4}{9}$ c. $\frac{11}{12}$ d. $\frac{11}{24}$

Solution:

Mean, x = x1 +x2 +x3 + ....... +xnn

= 14 +16 +34 +12 +512 +236
[Substitute the values.]

= 312 +212 +912 +612 +512 +8126

= 3312 × 6 = 1124

6.
The median of the observations $\frac{1}{4}$, $\frac{1}{6}$, $\frac{3}{4}$, $\frac{1}{2}$, $\frac{5}{12}$, $\frac{2}{3}$, $\frac{11}{12}$ is : a. $\frac{7}{12}$ b. $\frac{11}{12}$ c. $\frac{1}{2}$ d. $\frac{1}{6}$

Solution:

14, 16, 34, 12, 512, 23, 1112

312, 212, 912, 612, 512, 812, 1112
[Make the denominators same.]

212, 312, 512, 612, 812, 912, 1112
[Arrange in ascending order.]

Number of observations, n = 7
[n is odd.]

Location of Median is 7 + 12 = 4
[n + 12.]

Median is the 4th observation in the ordered list.

Median, M = 612 = 12

7.
The mean of the terms ($a$ + $b$), ($b$ + $c$) and ($c$ + $a$) is: a. $\mathrm{ab}$ + $\mathrm{bc}$ + $\mathrm{ca}$ b. $\frac{2}{3}$($a$ + $b$ + $c$) c. 2($a$ + $b$ + $c$) d. $a$ + $b$ + $c$

Solution:

x = x1 +x2 +x33

= a + b + b + c + c + a3 = 23(a + b + c)
[Substitute and Simplify.]

8.
The heights in cm of 9 students in the 9th grade are 149, 160, 151, 149, 151, 153, 160, 149, 158. Find the median. a. 149 b. 150 c. 152 d. 151

Solution:

149, 149, 149, 151, 151, 153, 158, 160, 160
[Arrange the data in increasing order.]

Number of observations, n = 9

Median is the 5th observation in the ordered list.
[n + 12 = 9 + 12 = 5.]

Median = 151

9.
The weight in kgs of 10 students are 35, 40, 33, 38, 32, 41, 45, 49, 38, and 42. What is the mean weight of the students in kg? a. 39 b. 49.5 c. 49 d. 39.3

Solution:

x = x1+x2+x3+ ... +xnn

= 35+40+33+38+32+41+45+49+38+4210
[Substitute the values.]

= 39310

= 39.3

The mean weight of 10 students is 39.3 kg.

10.
If the mean of the data 10, 15, $x$, 18, 12, 20 is 16, then the value of $x$ is : a. 31 b. 16 c. 14 d. 21

Solution:

x = x1+x2+x3+ ... +xnn

= 10 + 15 + x + 18 + 12 + 206
[Substitute the values.]

16 = 75 + x6
[Substitute: x = 16.]

96 = 75 + x

21 = x

The value of x is 21.