Mean of Ungrouped Frequency Distribution by Simple Method
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The mean of a data is defined as______
Let be three observations. The mean of these observations is:
Michael appeared in an examination and obtained the following marks out of 100.
NameMichaelPhysics87Chemistry82Biology82Mathematics98
Measure the central tendency of the above data.
Mean = 87.25, Median = 84.5 and Mode = 82
Mean = 86.25, Median = 82 and Mode = 98
Mean = 87.25, Median = 81 and Mode= 87
Mean= 87.25, Median = 87 and Mode = 82
The mean of 15 observations is 8. If the mean of the values of first eight observations is 7 and that of the values of last eight observations is 10, then what is the value of the 8th observation?
- 8
- 16
- 10
- 12
The distribution of the heights of 50 children are given. If the mean height for the distribution is 117.8 cm, then complete the following table.
Height110115x1120121125Number of students6814f143
x1=118, f1=15
x1=121, f1=15
x1=118, f1=8
x1=221, f1=45
- will become 5 times its original mean.
- will be decreased by 5
- will be increased by 5
- remains the same
Marks obtainedNumber of students0−10310−20420−301330−401540−505
- 28.75
- 25.25
- 26.44
- 29.91
- ¯¯¯¯¯X<¯¯¯¯¯¯¯X1
- ¯¯¯¯¯X>¯¯¯¯¯¯¯X2
- ¯¯¯¯¯X=¯¯¯¯¯¯¯X1−¯¯¯¯¯¯¯X22
- ¯¯¯¯¯¯¯X1<¯¯¯¯¯X<¯¯¯¯¯¯¯X2
80, 70, 72, 70, 36, 40, 36, 40, 92, 40, 50, 50, 56, 60, 70, 60, 60 and 88.
- 59.2
- 58
- 57.3
- 59.3
- 2
- 4
- 8
- 16
Find the value of 'p' if the mean of the following data is 17.
xi10p182125fi1015799
14
12
16
18
- 1400
- ₹1460
- ₹1480
- ₹1450
The mean marks of 100 students were found to be 40. Later on, it was discovered that a score of 53 was misread as 83. Then, what will be the correct Mean?
- 37.5
- 35.6
- 39.7
- 38.7
The following table shows the marks obtained by 5 students in a class room, find the average marks of the classroom.
MarksFrequency981651882541
77.6
78.6
78.9
76.6
xf4108111291613
- 10.33
- 10
- 43
- 444
The mean of the following distribution is:
xi10131619fi2576
15.55
16
15.2
15.35