Advertisements
Advertisements
प्रश्न
An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given in the following table:
Number of seats | 100 – 104 | 104 – 108 | 108 – 112 | 112 – 116 | 116 – 120 |
Frequency | 15 | 20 | 32 | 18 | 15 |
उत्तर
We first, find the class mark xi of each class and then proceed as follows.
Number of seats |
Class marks `(bb(x_i))` |
Frequency `(bb(f_i))` |
Deviation `bb(d_i = x_i - a)` |
`bb(f_i d_i)` |
100 – 104 | 102 | 15 | – 8 | – 120 |
104 – 108 | 106 | 20 | – 4 | – 80 |
108 – 112 | a = 110 | 32 | 0 | 0 |
112 – 116 | 114 | 18 | 4 | 72 |
116 – 120 | 118 | 15 | 8 | 120 |
`N = sumf_i = 100` | `sumf_i d_i = -8` |
∴ Assumed mean, a = 110
Class width, h = 4
And total observations, N = 100
By assumed mean method,
Mean `(barx) = a + (sumf_i d_i)/(sumf_i)`
= `110 + ((-8)/100)`
= 110 – 0.08
= 109.92
But seats cannot be in decimal, so number of seats is 109.
संबंधित प्रश्न
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 |
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
Number of mangoe | 50 − 52 | 53 − 55 | 56 − 58 | 59 − 61 | 62 − 64 |
Number of boxes | 15 | 110 | 135 | 115 | 25 |
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Find the value of p for the following distribution whose mean is 16.6
x | 8 | 12 | 15 | P | 20 | 25 | 30 |
f | 12 | 16 | 20 | 24 | 16 | 8 | 4 |
Find the mean of each of the following frequency distributions
Class interval | 0 - 8 | 8 - 16 | 16 - 24 | 24 - 32 | 32 - 40 |
Frequency | 6 | 7 | 10 | 8 | 9 |
For the following distribution, calculate mean using all suitable methods:
Size of item | 1 - 4 | 4 - 9 | 9 - 16 | 16 - 27 |
Frequency | 6 | 12 | 26 | 20 |
The following table shows the marks scored by 140 students in an examination of a certain paper:
Marks: | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 |
Number of students: | 20 | 24 | 40 | 36 | 20 |
Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.
The following distribution shows the daily pocket allowance given to the children of a multistorey building. The average pocket allowance is Rs 18.00. Find out the missing frequency.
Class interval | 11 - 13 | 13 - 15 | 15 - 17 | 17 - 19 | 19 - 21 | 21 - 23 | 23 - 25 |
Frequency | 7 | 6 | 9 | 13 | - | 5 | 4 |
If the mean of the following distribution is 27, find the value of p.
Class | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 |
Frequency | 8 | p | 12 | 13 | 10 |
If the mean of 25 observations is 27 and each observation is decreased by 7, what will be new mean?
Compute the mean for following data:
Class | 1-3 | 3-5 | 5-7 | 7-9 |
Frequency | 12 | 22 | 27 | 19 |
Find the mean of the following data, using assumed-mean method:
Class | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 | 100 - 120 |
Frequency | 20 | 35 | 52 | 44 | 38 | 31 |
The following table shows the age distribution of patients of malaria in a village during a particular month:
Age (in years) | 5 – 14 | 15 – 24 | 25 – 34 | 35 – 44 | 45 – 54 | 55 - 64 |
No. of cases | 6 | 11 | 21 | 23 | 14 | 5 |
Find the average age of the patients.
The following table shows the marks scored by 80 students in an examination:
Marks | 0 – 5 | 5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 | 35 – 40 |
No. of students |
3 | 10 | 25 | 49 | 65 | 73 | 78 | 80 |
Define mean.
If the mean of first n natural numbers is \[\frac{5n}{9}\], then n =
While computing mean of grouped data, we assume that the frequencies are ______.
Find the mean of the following distribution:
x | 4 | 6 | 9 | 10 | 15 |
f | 5 | 10 | 10 | 7 | 8 |
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 |
If the mean of the following distribution is 7.5, find the missing frequency ‘f’:
Variable : | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Frequency: | 20 | 17 | f | 10 | 8 | 6 | 7 | 6 |
Mean of 100 items is 49. It was discovered that three items which should have been 60, 70, 80 were wrongly read as 40, 20, 50 respectively. The correct mean is ______.
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?
In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula `barx = a + (sumf_i d_i)/(sumf_i)` where a is the assumed mean. a must be one of the mid-points of the classes. Is the last statement correct? Justify your answer.
Calculate the mean of the following data:
Class | 4 – 7 | 8 – 11 | 12 – 15 | 16 – 19 |
Frequency | 5 | 4 | 9 | 10 |
Find the mean, median and mode of the given data:
Class | 65 – 85 | 85 – 105 | 105 – 125 | 125 – 145 | 145 – 165 | 165 – 185 | 185 –205 |
Frequency | 8 | 7 | 22 | 17 | 13 | 5 | 3 |
The mean of the following frequency distribution is 25. Find the value of f.
Class | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
Frequency | 5 | 18 | 15 | f | 6 |
Which of the following cannot be determined graphically for a grouped frequency distribution?
Using step-deviation method, find mean for the following frequency distribution:
Class | 0 – 15 | 15 – 30 | 30 – 45 | 45 – 60 | 60 – 75 | 75 – 90 |
Frequency | 3 | 4 | 7 | 6 | 8 | 2 |