Free Statistics 03 Practice Test - 10th Grade
Question 1
The median for grouped data is formed by using the formula:
l+(n2−cff ) ×h
l−(n2−cff ) ×h.
l+(n2+cff ) ×h.
l+2n
SOLUTION
Solution : A
The median for grouped data is formed by l+(n2−cff ) ×h.
Where l is lower class limit of median class.
n is total number of observations.
cf is the cumulative frequency of the class preceding the median class.
f is the frequency of the median class and h is the class size.
Question 2
The following data shows monthly savings of 100 families . Calculate the mode of the given frequency distribution.
Monthly savings(Rs) Number of families 1000−2000142000−3000153000−4000214000−5000275000−600025
4790
4760
4750
4780
SOLUTION
Solution : C
Modal class is 4000-5000 within
f1 = 27, f0 = 21, f2 = 25, I = 4000, h = 1000
Mode = l + f1−f02f1−f0−f2×h
Mode = 4000 + 68×1000
= 4750
Question 3
Find the mode of the following data.
Class interval0−100100−200200−300300−400400−500Frequency7111598
SOLUTION
Solution :Modal Class is 200-300
Mode=I+(f1−f02f1−f0−f2)×h
=200+15−1130−11−9×100=240
Question 4
The below table shows that profit made by a group of shops in mall. Then the median is _____.
Profit per shop less than (%)102030405060No. of shops1230577794100
27.4
31.2
26.8
23.7
SOLUTION
Solution : A
Firstly, the cumulative frequency (CF) table is drawn.
C.ICFFrequency0−10121210−20301820−30572730−40772040−50941750−601006Total100
n2=1002=50
C.F nearest to 50 and greater than 50 is 57.
∴ Median=l+(n2−cff)×h
Where l is lower class limit of median class.
n is total number of observations.
cf is the cumulative frequency of the class preceding the median class.
f is the frequency of the median class and h is the class size.
Median=20+(50−3027)×10Median = 27.4
Question 5
The times (in seconds) taken by 150 athelets to run a 110 m hurdle race are tabulated below
Class13.8−1414−14.214.2−14.414.4−14.614.6−14.814.8−15Frequency245714820
The number of atheletes who completed the race in less than 14.6 s is
11
71
82
130
SOLUTION
Solution : C
The number of atheletes who completed the race in less than 14.6
= 2 + 4 + 5 + 71
= 82
Question 6
In the assumed mean method, if A is the assumed mean, then deviation di is :
SOLUTION
Solution : A
The deviation is di=xi−A
Question 7
In the assumed mean method, if A is the assumed mean, then deviation di is :
xi−A
xi+A
xi
A−xi
SOLUTION
Solution : A
In statistics, 'the assumed mean' is a method for calculating the arithmetic mean and standard deviation of a data set. It simplifies calculating accurate values by hand.
During the application of the short-cut method for finding the mean, the deviation d, are divisible by a common number ‘h’. In this case, the di = xi – A is reduced to a great extent as 'di' becomes dih. In this method, we divide the deviations by the same number to simplify calculations. The deviation is di=xi−A
Question 8
For which data set, the mean is not a good representative measures of central tendency.
{10, 12, 15, 17, 14, 15, 17}
{2, 20, 25, 17, 19, 22}
{5, 7, 5, 6, 4, 5, 7}
{17, 20, 25, 17, 19, 22}
SOLUTION
Solution : B
We know that extreme values in the data affect the mean. Here only for the data set {2, 20, 25, 17, 19, 22}, we find the extreme value as one class has the frequency as 2 and the others have frequency 20, 25, 17, 19 and 22.
Question 9
The following table gives the life time of 200 neon lamps. Find the median class.
SOLUTION
Solution : A
The median class is identified as the class whose cumulative frequency will be greater than n2, where 'n' is the sum of the frequencies. (n=200).
Next, we divide 200 by 2.
n2=2002=100
We now locate the class whose cumulative frequency is greater than 100.
"2500 - 3000" is the class whose cumulative frequency 155 is greater than 100.
∴The median class is 2500 - 3000.
Question 10
SOLUTION
Solution : B
Here we locate a class with the maximum frequency. From the given table we find the maximum frequency as 12 whose class is 2 - 3. Therefore the modal class is 2 - 3.