Advertisements
Advertisements
Question
The mean and standard deviation of marks obtained by 50 students of a class in three subjects, mathematics, physics and chemistry are given below:
| Subject | Mathematics | Physics | Chemistry |
| Mean | 42 | 32 | 40.9 |
| Standard Deviation | 12 | 15 | 20 |
Which of the three subjects shows the highest variability in marks and which shows the lowest?
Advertisements
Solution
We know: \[CV = \frac{\sigma}{\bar{X}} \times 100\]
\[\bar{{X_m}} = 42, \sigma_m = 12\]
\[ \bar{{X_p}} = 32, \sigma_p = 15\]
\[ \bar{{X_c}} = 40 . 9, \sigma_c = 20\]
CV of mathematics marks
Since CV of chemistry is the greatest, the variability of marks in chemistry is the highest and that of mathematics is the lowest.
APPEARS IN
RELATED QUESTIONS
Find the mean and variance for the data.
6, 7, 10, 12, 13, 4, 8, 12
Find the mean and variance for the first 10 multiples of 3.
Find the mean and variance for the data.
| xi | 6 | 10 | 14 | 18 | 24 | 28 | 30 |
| fi | 2 | 4 | 7 | 12 | 8 | 4 | 3 |
The following is the record of goals scored by team A in a football session:
|
No. of goals scored |
0 |
1 |
2 |
3 |
4 |
|
No. of matches |
1 |
9 |
7 |
5 |
3 |
For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?
The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.
Given that `barx` is the mean and σ2 is the variance of n observations x1, x2, …,xn. Prove that the mean and variance of the observations ax1, ax2, ax3, …,axn are `abarx` and a2 σ2, respectively (a ≠ 0).
The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:
|
Subject |
Mathematics |
Physics |
Chemistry |
|
Mean |
42 |
32 |
40.9 |
|
Standard deviation |
12 |
15 |
20 |
Which of the three subjects shows the highest variability in marks and which shows the lowest?
The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.
Find the mean, variance and standard deviation for the data:
2, 4, 5, 6, 8, 17.
Find the mean, variance and standard deviation for the data:
6, 7, 10, 12, 13, 4, 8, 12.
The variance of 15 observations is 4. If each observation is increased by 9, find the variance of the resulting observations.
The mean and standard deviation of 6 observations are 8 and 4 respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.
Find the standard deviation for the following data:
| x : | 3 | 8 | 13 | 18 | 23 |
| f : | 7 | 10 | 15 | 10 | 6 |
Calculate the standard deviation for the following data:
| Class: | 0-30 | 30-60 | 60-90 | 90-120 | 120-150 | 150-180 | 180-210 |
| Frequency: | 9 | 17 | 43 | 82 | 81 | 44 | 24 |
A student obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was later found that one observation was wrongly copied as 50, the correct figure being 40. Find the correct mean and S.D.
Calculate the mean, median and standard deviation of the following distribution:
| Class-interval: | 31-35 | 36-40 | 41-45 | 46-50 | 51-55 | 56-60 | 61-65 | 66-70 |
| Frequency: | 2 | 3 | 8 | 12 | 16 | 5 | 2 | 3 |
The means and standard deviations of heights ans weights of 50 students of a class are as follows:
| Weights | Heights | |
| Mean | 63.2 kg | 63.2 inch |
| Standard deviation | 5.6 kg | 11.5 inch |
Which shows more variability, heights or weights?
From the data given below state which group is more variable, G1 or G2?
| Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Group G1 | 9 | 17 | 32 | 33 | 40 | 10 | 9 |
| Group G2 | 10 | 20 | 30 | 25 | 43 | 15 | 7 |
If the sum of the squares of deviations for 10 observations taken from their mean is 2.5, then write the value of standard deviation.
The standard deviation of the data:
| x: | 1 | a | a2 | .... | an |
| f: | nC0 | nC1 | nC2 | .... | nCn |
is
If the standard deviation of a variable X is σ, then the standard deviation of variable \[\frac{a X + b}{c}\] is
If the S.D. of a set of observations is 8 and if each observation is divided by −2, the S.D. of the new set of observations will be
Let a, b, c, d, e be the observations with mean m and standard deviation s. The standard deviation of the observations a + k, b + k, c + k, d + k, e + k is
The standard deviation of first 10 natural numbers is
Let x1, x2, ..., xn be n observations. Let \[y_i = a x_i + b\] for i = 1, 2, 3, ..., n, where a and b are constants. If the mean of \[x_i 's\] is 48 and their standard deviation is 12, the mean of \[y_i 's\] is 55 and standard deviation of \[y_i 's\] is 15, the values of a and b are
The standard deviation of the observations 6, 5, 9, 13, 12, 8, 10 is
Life of bulbs produced by two factories A and B are given below:
| Length of life (in hours) |
Factory A (Number of bulbs) |
Factory B (Number of bulbs) |
| 550 – 650 | 10 | 8 |
| 650 – 750 | 22 | 60 |
| 750 – 850 | 52 | 24 |
| 850 – 950 | 20 | 16 |
| 950 – 1050 | 16 | 12 |
| 120 | 120 |
The bulbs of which factory are more consistent from the point of view of length of life?
Find the standard deviation of the first n natural numbers.
The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results:
Number of observations = 25, mean = 18.2 seconds, standard deviation = 3.25 seconds. Further, another set of 15 observations x1, x2, ..., x15, also in seconds, is now available and we have `sum_(i = 1)^15 x_i` = 279 and `sum_(i = 1)^15 x^2` = 5524. Calculate the standard derivation based on all 40 observations.
The mean and standard deviation of a set of n1 observations are `barx_1` and s1, respectively while the mean and standard deviation of another set of n2 observations are `barx_2` and s2, respectively. Show that the standard deviation of the combined set of (n1 + n2) observations is given by
S.D. = `sqrt((n_1(s_1)^2 + n_2(s_2)^2)/(n_1 + n_2) + (n_1n_2 (barx_1 - barx_2)^2)/(n_1 + n_2)^2)`
If for distribution `sum(x - 5)` = 3, `sum(x - 5)^2` = 43 and total number of items is 18. Find the mean and standard deviation.
If the variance of a data is 121, then the standard deviation of the data is ______.
The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.
