Solutions: Statistics - Test 1

Q1. (c) 0

The sum of deviations of observations from their mean is always zero. $\sum (x_i - \bar{x}) = 0$.

Q2. (b) centred at the classmarks of the classes

In grouped data, we assume the frequency of each class interval is concentrated at its mid-point (class mark).

Q3. (b) 8

Mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.
$\sum f_i = 4+5+y+1+2 = 12+y$.
$\sum f_i x_i = 4(1) + 5(2) + y(3) + 1(4) + 2(5) = 4 + 10 + 3y + 4 + 10 = 28 + 3y$.
$2.6 = \frac{28+3y}{12+y} \Rightarrow 2.6(12+y) = 28+3y \Rightarrow 31.2 + 2.6y = 28 + 3y$.
$3.2 = 0.4y \Rightarrow y = 8$.

Q4. (a) Both A and R are true and R is the correct explanation of A.

Sum of first $n$ odd natural numbers is $n^2$.
Mean = $\frac{n^2}{n} = n$. Assertion is true. Reason is the definition of mean, which explains the calculation.

Q5. First 5 prime numbers: 2, 3, 5, 7, 11.

Sum = $2+3+5+7+11 = 28$.
Mean = $\frac{28}{5} = 5.6$.

Q6. Sum of 5 numbers = $5 \times 18 = 90$.

Sum of 4 numbers (after excluding one) = $4 \times 16 = 64$.
Excluded number = $90 - 64 = 26$.

Q7. Value of $\sum (f_i x_i - \bar{x})$.

$\sum (f_i x_i - \bar{x}) = \sum f_i x_i - \sum f_i \bar{x} = \sum f_i x_i - \bar{x} \sum f_i$.
Since $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$, we have $\sum f_i x_i = \bar{x} \sum f_i$.
Therefore, expression = $\bar{x} \sum f_i - \bar{x} \sum f_i = 0$.

Q8. Mean Literacy Rate.

Class Marks ($x_i$): 50, 60, 70, 80, 90.
$f_i$: 3, 10, 11, 8, 3. $\sum f_i = 35$.
$f_i x_i$: 150, 600, 770, 640, 270.
$\sum f_i x_i = 2430$.
Mean = $\frac{2430}{35} = 69.43\%$.

Q9. Missing frequency $f$. Mean = 18.

$x_i$: 12, 14, 16, 18, 20, 22, 24.
$f_i$: 7, 6, 9, 13, $f$, 5, 4. $\sum f_i = 44 + f$.
$f_i x_i$: 84, 84, 144, 234, $20f$, 110, 96. $\sum f_i x_i = 752 + 20f$.
Mean $18 = \frac{752 + 20f}{44 + f} \Rightarrow 18(44+f) = 752 + 20f$.
$792 + 18f = 752 + 20f \Rightarrow 2f = 40 \Rightarrow f = 20$.

Q10. Step Deviation Method. Assumed Mean $a = 125$, $h = 50$.

Classes: 0-50, 50-100, 100-150, 150-200, 200-250, 250-300.
$x_i$: 25, 75, 125, 175, 225, 275.
$u_i = (x_i - 125)/50$: -2, -1, 0, 1, 2, 3.
$f_i$: 17, 35, 43, 40, 21, 24. $\sum f_i = 180$.
$f_i u_i$: -34, -35, 0, 40, 42, 72. $\sum f_i u_i = 85$.
Mean $\bar{x} = a + (\frac{\sum f_i u_i}{\sum f_i}) \times h = 125 + (\frac{85}{180}) \times 50$.
$= 125 + \frac{85}{18} \times 5 = 125 + \frac{425}{18} = 125 + 23.61 = 148.61$.

(i) Class mark of 14-16 = $\frac{14+16}{2} = 15$.

(ii) Mean Mileage: $x_i$: 11, 13, 15, 17, 19. $f_i$: 7, 12, 18, 8, 5. $\sum f = 50$.
$f_i x_i$: 77, 156, 270, 136, 95. $\sum f x = 734$.
Mean = $\frac{734}{50} = 14.68$ km/l.

(iii) If each observation is increased by $k$, the mean increases by $k$.
New Mean = $14.68 + 2 = 16.68$ km/l.