Advertisements
Advertisements
Question
Let x1, x2, x3, x4, x5 be the observations with mean m and standard deviation s. The standard deviation of the observations kx1, kx2, kx3, kx4, kx5 is ______.
Options
k + s
`s/k`
ks
s
Solution
Let x1, x2, x3, x4, x5 be the observations with mean m and standard deviation s. The standard deviation of the observations kx1, kx2, kx3, kx4, kx5 is ks.
Explanation:
Here m = `(sumx_i)/n`
s = `sqrt((sumx_i^2)/5 - ((sumx_i)/5)^2`
∴ S.D. = `sqrt((k^2sumx_i^2)/5 - ((ksumx_i)/5)^2`
= `sqrt((k^2sumx_i^2)/5 - k^2 ((sumx_i)/5)^2`
= `ksqrt((sumx_i^2)/5 - ((sumx_i)/5)^2`
= ks
APPEARS IN
RELATED QUESTIONS
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 sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:
`sum_(i-1)^50 x_i = 212, sum_(i=1)^50 x_i^2 = 902.8, sum_(i=1)^50 y_i = 261, sum_(i = 1)^50 y_i^2 = 1457.6`
Which is more varying, the length or weight?
The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12 and 14. 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 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
- If wrong item is omitted.
- If it is replaced by 12.
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 15, 22, 27, 11, 9, 21, 14, 9.
The variance of 15 observations is 4. If each observation is increased by 9, find the variance of the resulting observations.
For a group of 200 candidates, the mean and standard deviations of scores were found to be 40 and 15 respectively. Later on it was discovered that the scores of 43 and 35 were misread as 34 and 53 respectively. Find the correct mean and standard deviation.
The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation?
The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
(i) If wrong item is omitted
(ii) if it is replaced by 12.
Find the standard deviation for the following distribution:
x : | 4.5 | 14.5 | 24.5 | 34.5 | 44.5 | 54.5 | 64.5 |
f : | 1 | 5 | 12 | 22 | 17 | 9 | 4 |
Find the mean and variance of frequency distribution given below:
xi: | 1 ≤ x < 3 | 3 ≤ x < 5 | 5 ≤ x < 7 | 7 ≤ x < 10 |
fi: | 6 | 4 | 5 | 1 |
Mean and standard deviation of 100 observations were found to be 40 and 10 respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.
Two plants A and B of a factory show following results about the number of workers and the wages paid to them
Plant A | Plant B | |
No. of workers | 5000 | 6000 |
Average monthly wages | Rs 2500 | Rs 2500 |
Variance of distribution of wages | 81 | 100 |
In which plant A or B is there greater variability in individual wages?
Find the coefficient of variation for the following data:
Size (in cms): | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 |
No. of items: | 2 | 8 | 20 | 35 | 20 | 15 |
If X and Y are two variates connected by the relation
The standard deviation of the data:
x: | 1 | a | a2 | .... | an |
f: | nC0 | nC1 | nC2 | .... | nCn |
is
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
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
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?
A set of n values x1, x2, ..., xn has standard deviation 6. The standard deviation of n values x1 + k, x2 + k, ..., xn + k will be ______.
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.