Free Measures of Dispersion 01 Practice Test - 11th Grade - Commerce
Question 1
Find the range of the following data
92629−1062273424123133023413−95−6−4
SOLUTION
Solution : D
Range is the difference between the largest and the smallest observation.
Range=34−(−10)=44
Question 2
Steven Gerrard, arguably one of the most consistent and skilled players in the English premier league was given the following monthly ratings by a football website for the past 40 months.
Rating688.59710Number of months with the rating2810785
Calculate how his rating is deviating from his average performance in terms of mean deviation about mean.
0.825
0.85
0.9
1
SOLUTION
Solution : A
Mean=∑FX∑F
Mean, ¯X=6(2)+8(8)+(8.5)(10)+9(7)+7(8)+10(5)2+8+10+7+8+5=33040=8.25
The deviations from the mean are tabulated below.
XFD=|X−¯X|FD622.254.5880.2528.5100.252.5970.755.25781.25101051.758.75 ∑F=40 ∑FD=33M.D(¯X)=∑FD∑F=3340=0.825
∴ Mean deviation about mean for his performance across 40 months about his average performance rating of 8.25 is 0.825 which shows he is quite consistent.
Question 3
The heights of six students in a class are 169 cm, 172 cm, 178 cm, 170 cm, 166 cm and 174 cm. Find the standard deviation of heights using the assumed mean method.
SOLUTION
Solution : C
Let 170 be the assumed mean (A).
XD=X−AD2169−111722417886417000166−416174416 ∑D=9∑D2=101
σ=√∑D2N−(∑DN)2=√1016−(96)2=√14.58=3.82
Question 4
The mark distribution of 40 students in an exam is shown below. Find the standard deviation of the distribution.
Marks0−1010−2020−3030−4040−50No. of Students7101265
10.28
12.49
14.76
15.83
SOLUTION
Solution : B
XFFXD=X−¯XD2FD25735−1832422681510150−8646402512300244835621012144864455225224842420∑F=40∑FX=920∑FD2=6240
Mean, ¯X=∑FX∑F=92040=23
σ=√∑FD2∑F=√624040=√156=12.49
Question 5
The number of years of education acquired by a sample of 100 adults in a locality is as follows.
Years of education0−44−88−1212−1616−2020−24No. of people1022322484
2.85
3.75
4.95
6.05
SOLUTION
Solution : C
Let the assumed mean be 10 i.e. A = 10
XFD=X−AFDFD2210−8−80640622−4−88352103200014244963841888645122241248576∑F=100∑FD=40∑FD2=2464
σ=√∑FD2∑F−(∑FD∑F)2=√2464100−(40100)2=√24.64−0.16=√24.48=4.95
Question 6
The average ambient temperatures (in degrees celsius) for a region for 15 years from 2001-2015 is given below.
26.3, 29.7, 31.0, 29.8, 26.8, 27.6, 27.4, 27.0, 27.6, 27.8, 30.0, 32.0, 28.9, 30.5, 29.5
Calculate the coefficient of range for the temperatures.
SOLUTION
Solution : D
The lowest temperature is 26.3 degrees and the highest temperature is 32.0 degrees.
Coefficient of range=L−SL+S=32∘C−26.3∘C32∘C+26.3∘C=5.7∘C58.3∘C=0.098
Question 7
Data A has a mean of 100 and standard deviation of 10. Data B has a mean of 1000 and standard deviation of 50. Which data shows more variability?
SOLUTION
Solution : A
To compare dissimilar data, relative measures should be used. Here, the coefficient of variation of set A and set B can be compared.
COVA=10100×100=10%
COVB=501000×100=5%
Hence, data A shows more variability.
Question 8
Given below is the Lorenz curve representing the percentage revenues of car manufacturing companies worldwide.
The top 20 percentage of the companies generate what percentage of revenue?
SOLUTION
Solution : D
From the Lorenz curve, the bottom 80% of the companies generate 40% of the revenue. Hence, the top 20% of the companies generate 60% of the revenue.
Question 9
For a data set, the coefficient of range is 0.5. The largest observation is 12. Find the smallest observation.
SOLUTION
Solution : C
Given that the coefficient of range is 0.5.
L−SL+S=0.5 ⇒L=3S
Given that L=12.
∴ 3S=12S=4
Question 10
The coefficient of variation of data pertaining to the goals scored by two teams A and B in a football season are 0.35 and 0.72 respectively. The true statement is
SOLUTION
Solution : C
The coefficient of variation is a measure of consistency. Lower, the C.V, more consistent the data is. In this case, team A has a lower C.V and hence, is more consistent.