Exercise 14.2: Statistics — NCERT Solutions

The first ten prime numbers are: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29.$

Sum $= 2+3+5+7+11+13+17+19+23+29 = 129.$

Number of observations $n = 10$.

$\bar{x} = \frac{\text{Sum of observations}}{n} = \frac{129}{10} = 12.9.$

The mean of the first ten prime numbers is $12.9$.

Mean $= \dfrac{\text{sum}}{n}$, so sum of 8 observations: $\text{Sum} = 15 \times 8 = 120.$

Remove the observation 20: $\text{New sum} = 120 - 20 = 100,$ with $n = 7$ remaining observations.

$\text{New mean} = \frac{100}{7} \approx 14.29.$

Original sum $= 27 \times 5 = 135.$

When each of the 5 numbers is increased by 4, the total increases by $4 \times 5 = 20$: $\text{New sum} = 135 + 20 = 155.$

$\text{New mean} = \frac{155}{5} = 31.$

Shortcut: adding a constant $c$ to every observation increases the mean by $c$: $27 + 4 = 31$.

Use $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$.

| Goals $x_i$ | Frequency $f_i$ | $f_i x_i$ | |:---:|:---:|:---:| | 0 | 3 | 0 | | 1 | 6 | 6 | | 2 | 5 | 10 | | 3 | 4 | 12 | | 4 | 2 | 8 | | Total | 20 | 36 |

$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{36}{20} = 1.8 \text{ goals per match}.$

Mean $= \dfrac{\text{sum}}{6} = 52$, so the total must be: $\text{Sum} = 52 \times 6 = 312.$

Add the known marks: $45 + 50 + 55 + x + 60 + 40 = 312$ $250 + x = 312$ $x = 312 - 250 = 62.$

So the missing mark is $x = 62$.

The class mark is $x_i = \dfrac{\text{lower limit} + \text{upper limit}}{2}$.

| Class | Class mark $x_i$ | $f_i$ | $f_i x_i$ | |:---:|:---:|:---:|:---:| | 150–155 | 152.5 | 4 | 610 | | 155–160 | 157.5 | 6 | 945 | | 160–165 | 162.5 | 7 | 1137.5 | | 165–170 | 167.5 | 3 | 502.5 | | Total | | 20 | 3195 |

$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{3195}{20} = 159.75 \text{ cm}.$