Advertisements
Advertisements
प्रश्न
If \[u_i = \frac{x_i - 25}{10}, \Sigma f_i u_i = 20, \Sigma f_i = 100, \text { then }\]`overlineX`
पर्याय
23
24
27
25
उत्तर
Given:
\[u_i = \frac{x_i - 25}{10}, \Sigma f_i u_i = 20 \text { and } \Sigma f_i = 100\]
Now, `u_i = (x_i - A)/h`= \[\frac{x_i - 25}{10}\]
Therefore, h = 10 and A = 25
We know that
`overlineX = A+h{1/N sum f_iu_i}`
`= 25 + 10{1/100 xx 20}`
`=25 +10 xx 1/5`
`=25 + 2`
`overlineX = 27`
APPEARS IN
संबंधित प्रश्न
The given distribution shows the number of wickets taken by the bowlers in one-day international cricket matches:
| Number of Wickets | Less than 15 | Less than 30 | Less than 45 | Less than 60 | Less than 75 | Less than 90 | Less than 105 | Less than 120 |
| Number of bowlers | 2 | 5 | 9 | 17 | 39 | 54 | 70 | 80 |
Draw a ‘less than type’ ogive from the above data. Find the median.
The heights of 50 girls of Class X of a school are recorded as follows:
| Height (in cm) | 135 - 140 | 140 – 145 | 145 – 150 | 150 – 155 | 155 – 160 | 160 – 165 |
| No of Students | 5 | 8 | 9 | 12 | 14 | 2 |
Draw a ‘more than type’ ogive for the above data.
What is the cumulative frequency of the modal class of the following distribution?
| Class | 3 – 6 | 6 – 9 | 9 – 12 | 12 – 15 | 15 – 18 | 18 – 21 | 21 – 24 |
|
Frequency |
7 | 13 | 10 | 23 | 54 | 21 | 16 |
The following table, construct the frequency distribution of the percentage of marks obtained by 2300 students in a competitive examination.
| Marks obtained (in percent) | 11 – 20 | 21 – 30 | 31 – 40 | 41 – 50 | 51 – 60 | 61 – 70 | 71 – 80 |
| Number of Students | 141 | 221 | 439 | 529 | 495 | 322 | 153 |
(a) Convert the given frequency distribution into the continuous form.
(b) Find the median class and write its class mark.
(c) Find the modal class and write its cumulative frequency.
The following is the cumulative frequency distribution ( of less than type ) of 1000 persons each of age 20 years and above . Determine the mean age .
| Age below (in years): | 30 | 40 | 50 | 60 | 70 | 80 |
| Number of persons : | 100 | 220 | 350 | 750 | 950 | 1000 |
The mean of a discrete frequency distribution xi / fi, i = 1, 2, ......, n is given by
Consider the following frequency distribution :
| Class: | 0-5 | 6-11 | 12-17 | 18-23 | 24-29 |
| Frequency: | 13 | 10 | 15 | 8 | 11 |
The upper limit of the median class is
The marks obtained by 100 students of a class in an examination are given below.
| Mark | No. of Student |
| 0 - 5 | 2 |
| 5 - 10 | 5 |
| 10 - 15 | 6 |
| 15 - 20 | 8 |
| 20 - 25 | 10 |
| 25 - 30 | 25 |
| 30 - 35 | 20 |
| 35 - 40 | 18 |
| 40 - 45 | 4 |
| 45 - 50 | 2 |
Draw 'a less than' type cumulative frequency curves (ogive). Hence find the median.
Change the following distribution to a 'more than type' distribution. Hence draw the 'more than type' ogive for this distribution.
| Class interval: | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 | 80−90 |
| Frequency: | 10 | 8 | 12 | 24 | 6 | 25 | 15 |
Consider the following distribution:
| Marks obtained | Number of students |
| More than or equal to 0 | 63 |
| More than or equal to 10 | 58 |
| More than or equal to 20 | 55 |
| More than or equal to 30 | 51 |
| More than or equal to 40 | 48 |
| More than or equal to 50 | 42 |
The frequency of the class 30 – 40 is:
