Advertisements
Advertisements
Question
If the mean of the following frequency distribution is 18, find the missing frequency.
Class interval | 11 – 13 | 13 – 15 | 15 – 17 | 17 – 19 | 19 – 21 | 21 – 23 | 23 – 25 |
Frequency | 3 | 6 | 9 | 13 | f | 5 | 4 |
Solution
Class | Mid point `(bb(x_i))` |
Frequency `(bb(f_i))` |
`bb(f_ix_i)` |
11 – 13 | 12 | 3 | 36 |
13 – 15 | 14 | 6 | 84 |
15 – 17 | 16 | 9 | 144 |
17 – 19 | 18 | 13 | 234 |
19 – 21 | 20 | f | 20f |
21 – 23 | 22 | 5 | 110 |
23 – 25 | 24 | 4 | 96 |
`sumf_i = 40 + f` | `sumf_ix_i = 704 + 20f` |
Mean = `(sumf_ix_i)/(sumf_i)`
∴ 18 = `(704 + 20f)/(40 + f)` ...[∵ Given, mean = 18]
⇒ 18(40 + f) = 704 + 20f
⇒ 720 + 18f = 704 + 20f
⇒ 2f = 16
⇒ f = 8
So, missing frequency f is 8.
APPEARS IN
RELATED QUESTIONS
The following table gives the frequency distribution of trees planted by different Housing Societies in a particular locality:
No. of Trees | No. of Housing Societies |
10-15 | 2 |
15-20 | 7 |
20-25 | 9 |
25-30 | 8 |
30-35 | 6 |
35-40 | 4 |
Find the mean number of trees planted by Housing Societies by using ‘Assumed Means Method’
Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarized as follows. Fine the mean heartbeats per minute for these women, choosing a suitable method.
Number of heartbeats per minute | 65 - 68 | 68 - 71 | 71 - 74 | 74 - 77 | 77 - 80 | 80 - 83 | 83 - 86 |
Number of women | 2 | 4 | 3 | 8 | 7 | 4 | 2 |
Find the mean of each of the following frequency distributions
Classes | 25 - 29 | 30 - 34 | 35 - 39 | 40 - 44 | 45 - 49 | 50 - 54 | 55 - 59 |
Frequency | 14 | 22 | 16 | 6 | 5 | 3 | 4 |
The mean of n observation is `overlineX` .f the first item is increased by 1, second by 2 and so on, then the new mean is
If the mean of the following distribution is 2.6, then the value of y is:
Variable (x) | 1 | 2 | 3 | 4 | 5 |
Frequency | 4 | 5 | y | 1 | 2 |
The mean of n observation is `overlineX`. If the first observation is increased by 1, the second by 2, the third by 3, and so on, then the new mean is
Mean of a certain number of observation is `overlineX`. If each observation is divided by m(m ≠ 0) and increased by n, then the mean of new observation is
The mean of the following distribution is 6. Find the value at P:
x | 2 | 4 | 6 | 10 | P + 5 |
f | 3 | 2 | 3 | 1 | 2 |
The average weight of a group of 25 men was calculated to be 78.4 kg. It was discovered later that one weight was wrongly entered as 69 kg instead of 96 kg. What is the correct average?
Calculate the mean of the following data:
Class | 4 – 7 | 8 – 11 | 12 – 15 | 16 – 19 |
Frequency | 5 | 4 | 9 | 10 |