10th Grade > Mathematics
STATISTICS MCQs
:
A
¯x=A+∑fidi∑fi
In the formula, the A stands for the assumed mean.
:
B
The given table can be written as:
Marks0−1010−2020−3030−4040−50 No. of students551389 Cumulative Frequency510233140
No. of students can be calculated as follows:
5
10 - 5 = 5
23 - 10 = 13
31 - 23 = 8
40 - 31 = 9
∴n2=402=20
Since there are a total of 40 observations, the median class is the one whose cumulative frequency is closest to and greater than 20.
Here the cumulative frequency for the class interval '10 - 20' is 23 which is greater than 20.
The class interval 20 - 30 is the required answer.
:
A
The given table can be written as:
Height (in cm) Less than 140 Less than 145 Less than 150Less than 155Less than 160Less than 165No. of girls41129404651
Height (in cm) Less than 140140−145145−150150−155155−160160−165Cumulative Frequency41129404651
n = 51
∴n2=512
= 25.5
Since there are a total of 51 observations, the median class is the one whose cumulative frequency is closest to and greater than 25.5. The cumulative frequency for the class '145 - 150' is 29 which is greater than 25.5. The median class is '140 - 145'. Thus, the class interval '145 - 150' is the required answer.
:
A
ClassIntervalFrequencyCumulativeFrequency60−702270−803580−9051090−1001626100−1101440110−1201353120−130760
n2=30.
So, the median class is the one whose cumulative frequency is closest to and greater than 30.
∴ the median class is 100-110
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=100+30−2614×10
∴ Median=102.86
:
The method in question is the step-deviation method.
:
A
For grouped data, using the direct method first, we find the sum of the values of all the observations after multiplying them with their respective frequencies. Then we divide this result by the total number of observations (sum of all frequencies).
:
C
Mean = 45 and Mode = 60
3 Median = Mode +2 Mean
3 Median = 60 +2(45)
3 Median = 150
Median=50
:
B
Using, Mode = l+f1−f02f1−f0−f2×h
where,
Maximum class frequency, f1=8
The class corresponding to the frequency =20−40
Lower limit of the modal class, l=20
Frequency of class preceding the modal class, f0=7
Frequency of class succeeding the modal class, f2=2
Now,Mode=20+8−716−7−2×20
Mode=20+17×20
∴Mode=22.86
:
B
The empirical relationship between Median, Mode and Mean is Mode = 3 Median - 2 Mean
:
A
l+(f1−f02f1−f0−f2) ×h is used to determine the mode .
A mode is that value among the observations which occurs most often, that is, the value of the observation having the maximum frequency.