Chapter 13: Statistics

We build the following table from the given data:

Here $n=8$ and $\displaystyle\sum_{i=1}^8 x_i=80$.

Applying the mean deviation formula about the mean:

Here $n =$ number of items $= 10$. First, we compute the mean:

Next, we build the deviation table:

The mean deviation from the mean is:

We first arrange the data in ascending order:

Since $n = 12$ (even), the median is the average of the $\dfrac{n}{2} = 6$th and $\dfrac{n}{2}+1 = 7$th terms:

We now build the deviation table:

Applying the mean deviation formula about median:

Arranging the data in ascending order:

Since $n = 10$ (even), the median is the average of the 5th and 6th terms:

Setting up the deviation table:

We build a comprehensive working table from the given data:

Here $N = \displaystyle\sum_{i=1}^5 f_i = 25$ and $\displaystyle\sum_{i=1}^5 f_i x_i = 350$.

We construct the working table:

Here $N = 80$ and $\displaystyle\sum_{i=1}^5 f_i x_i = 4000$.

The observations are already in ascending order. We form the cumulative frequency table:

Here $N = 26$ (even). The median is the average of the 13th and 14th observations. Since all observations from the 9th to the 14th have $x = 7$, we get:

The observations are in ascending order. We form the cumulative frequency table:

Here $N = 29$ (odd), so the median is the value of $x$ at the $\left(\dfrac{29+1}{2}\right)$th = 15th observation. From the c.f. column, this lies in the group with $x = 30$.

We take assumed mean $a = 350$ and $h = 100$, then build the step-deviation table:

Here $N = 50$, $\displaystyle\sum f_i u_i = 4$.

We take assumed mean $a = 120$ and $h = 10$, then form the step-deviation table:

Here $N = 100$, $\displaystyle\sum f_i y_i = 53$.

We form the following cumulative frequency table:

Here $N = 50$. The $\dfrac{N}{2} = 25$th item falls in the class $20$–$30$ (c.f. just $\geq 25$ is 28).

So the median class is $20$–$30$ with $l = 20$, $f = 14$, $C = 14$, $h = 10$:

The class intervals are discontinuous (inclusive), so we first convert them to continuous form by subtracting 0.5 from each lower limit and adding 0.5 to each upper limit:

Since $\dfrac{N}{2} = 50$ and the c.f. just $\geq 50$ is 63, the median class is $35.5$–$40.5$.

Here $l = 35.5$, $f = 26$, $C = 37$, $h = 5$:

Computing mean deviation:

We build the deviation table:

Here $n = 8$ and $\displaystyle\sum_{i=1}^8 x_i = 72$.

The observations are $x: 1, 2, 3, \ldots, n$.

Using the variance formula $\sigma^2 = \dfrac{1}{n}\Sigma x^2 – \bar{x}^2$:

The first 10 multiples of 3 are: $3, 6, 9, 12, 15, 18, 21, 24, 27, 30$.

Since the mean is not a natural number, we use the step-deviation method with assumed mean $A = 18$:

We take assumed mean $A = 18$ and form the following table:

Here $N = 40$, $\displaystyle\sum f_i d_i = 40$, $\displaystyle\sum f_i d_i^2 = 1776$.

We take assumed mean $A = 100$ and form the working table:

Here $N = 22$, $\displaystyle\sum f_i d_i = 0$.

We take assumed mean $A = 64$ and form the step-deviation table:

Here $N = 100$, $\displaystyle\sum f_i d_i = 0$, $\displaystyle\sum f_i d_i^2 = 286$.

We take $A = 105$ and $h = 30$, then form the step-deviation table:

We take assumed mean $A = 25$ and $h = 10$, then form the working table:

We take assumed mean $A = 92.5$ and $h = 5$, then form the table:

We first make the data continuous: classes become $32.5$–$36.5$, $36.5$–$40.5$, etc. Taking $A = 42.5$ and $h = 4$:

For Group A (taking $A = 45$, $h = 10$):

For Group B (same $A = 45$, $h = 10$):

Since both groups have the same mean, the group with larger standard deviation is more variable. Since $\sigma_y > \sigma_x$, Group B is more variable.

For Shares X (assumed mean $A = 50$):

For Shares Y (assumed mean $A = 104$):

Comparing Coefficients of Variation:

Since C.V. for Y is much smaller than C.V. for X, Share Y is more stable.

(i) We are given $\bar{x} = \bar{y} = ₹5253$ (same for both firms), $n_1 = 586$ (Firm A), $n_2 = 648$ (Firm B).

Total monthly wages paid = No. of workers × Mean wage.

Since $\bar{x} = \bar{y}$ and $n_2 > n_1$, Firm B pays a larger total amount as monthly wages.

(ii) We are given $\sigma_A^2 = 100$ and $\sigma_B^2 = 121$, giving $\sigma_A = 10$ and $\sigma_B = 11$.

Since both firms have the same mean and $\sigma_B > \sigma_A$, Firm B shows greater variability in individual wages.

For Team A:

For Team B: Mean $\bar{y} = 2$, S.D. $= 1.25$, so Variance $\sigma_B^2 = (1.25)^2 = 1.5625$.

Since $\bar{x} = \bar{y} = 2$ and $\sigma_A^2 < \sigma_B^2$, Team A has smaller variability. Therefore, Team A is more consistent.

For Length $(x)$:

For Weight $(y)$:

Since C.V. for $y >$ C.V. for $x$, weight is more varying than length.

Let the two unknown observations be $a$ and $b$. Using $\bar{x} = 9$ and $n = 8$:

Using the variance formula $\sigma^2 = \dfrac{1}{n}\Sigma x^2 – \bar{x}^2$ with $\sigma^2 = 9.25$:

From (i): $b = 12-a$. Substituting into (ii):

When $a = 4$: $b = 8$. When $a = 8$: $b = 4$.

∴ The remaining two observations are $\boxed{4 \text{ and } 8}$.

Let the two unknown observations be $a$ and $b$. Using $\bar{x} = 8$ and $n = 7$:

Using $\sigma^2 = 16$ with the variance formula:

From (i): $b = 14-a$. Substituting into (ii):

When $a = 6$: $b = 8$. When $a = 8$: $b = 6$.

∴ The remaining two observations are $\boxed{6 \text{ and } 8}$.

Let the six observations be $x_1, x_2, \ldots, x_6$.

When each observation is multiplied by 3, the new observations are $y_i = 3x_i$.

We are given:

When each observation is multiplied by $a (\neq 0)$, the new observations are $ax_1, ax_2, \ldots, ax_n$.

New Mean:

∴ New mean $= a\bar{x}$.   (Proved)

New Variance:

∴ New variance $= a^2\sigma^2$.   (Proved)

Here $n = 20$, $\bar{x} = 10$, $\sigma = 2$.

Using the formula $\dfrac{1}{n}\Sigma x_i^2 – \bar{x}^2 = \sigma^2 = 4$:

(i) Wrong item 8 is omitted:

(ii) Wrong entry 8 replaced by 12:

The Coefficient of Variation (C.V.) measures relative variability. We calculate it for each subject using $\text{C.V.} = \dfrac{\text{S.D.}}{\text{Mean}} \times 100$:

Since Chemistry has the greatest C.V., it shows the highest variability.

Since Mathematics has the smallest C.V., it shows the lowest variability.

With $n = 100$, $\bar{x} = 20$, $\sigma = 3$:

Omitting the three incorrect observations $21, 21, 18$:

Using $\sigma^2 = 9$ to find $\Sigma x_i^2$: