Advertisements
Advertisements
प्रश्न
If the mean of 6, 7, x, 8, y, 14 is 9, then ______.
विकल्प
x + y = 21
x + y = 19
x − y = 19
x − y = 21
उत्तर
If the mean of 6, 7, x, 8, y, 14 is 9, then x + y = 19.
Explanation:
The given observations are 6, 7, x, 8, y, 14.
Mean = 9 ...(Given)
\[\Rightarrow \frac{6 + 7 + x + 8 + y + 14}{6} = 9\]
\[\Rightarrow 35 + x + y = 54\]
\[\Rightarrow x + y = 54 - 35 = 19\]
APPEARS IN
संबंधित प्रश्न
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs.18. Find the missing frequency f.
Daily pocket allowance (in Rs | 11 - 13 | 13 - 15 | 15 - 17 | 17 - 19 | 19 - 21 | 21 - 23 | 23 - 25 |
Number of workers | 7 | 6 | 9 | 13 | f | 5 | 4 |
If the mean of the following data is 20.6. Find the value of p.
x | 10 | 15 | P | 25 | 35 |
f | 3 | 10 | 25 | 7 | 5 |
If the mean of the following data is 18.75. Find the value of p.
x | 10 | 15 | P | 25 | 30 |
f | 5 | 10 | 7 | 8 | 2 |
The following table gives the distribution of total household expenditure (in rupees) of manual workers in a city. Find the average expenditure (in rupees) per household.
Expenditure (in rupees) (x1) |
Frequency(f1) |
100 - 150 | 24 |
150 - 200 | 40 |
200 - 250 | 33 |
250 - 300 | 28 |
300 - 350 | 30 |
350 - 400 | 22 |
400 - 450 | 16 |
450 - 500 | 7 |
A school has 4 sections of Chemistry in class X having 40, 35, 45 and 42 students. The mean marks obtained in Chemistry test are 50, 60, 55 and 45 respectively for the 4 sections. Determine the overall average of marks per student.
The median from the table is?
Value | Frequency |
7 | 2 |
8 | 1 |
9 | 4 |
10 | 5 |
11 | 6 |
12 | 1 |
13 | 3 |
There is a grouped data distribution for which mean is to be found by step deviation method.
Class interval | Number of Frequency (fi) | Class mark (xi) | di = xi - a | `u_i=d_i/h` |
0 - 100 | 40 | 50 | -200 | D |
100 - 200 | 39 | 150 | B | E |
200 - 300 | 34 | 250 | 0 | 0 |
300 - 400 | 30 | 350 | 100 | 1 |
400 - 500 | 45 | 450 | C | F |
Total | `A=sumf_i=....` |
Find the value of A, B, C, D, E and F respectively.
(For an arranged data) The median is the ______.
If xi’s are the midpoints of the class intervals of grouped data, fi’s are the corresponding frequencies and `barx` is the mean, then `sum(f_ix_i - barx)` is equal to ______.
The distribution given below shows the runs scored by batsmen in one-day cricket matches. Find the mean number of runs.
Runs scored |
0 – 40 | 40 – 80 | 80 – 120 | 120 – 160 | 160 – 200 |
Number of batsmen |
12 | 20 | 35 | 30 | 23 |