C7 Pg-212 Mathematics

Question 17

Given frequency distribution:

\(x_i\)	22	23	24	25	26	27	28
\(f_i\)	5	12	9	\(x\)	20	14	22

Mean of the data = \(25.78\) Formula: \[ \bar{x}=\frac{\sum f_i x_i}{\sum f_i} \] Sum of frequencies: \[ \sum f_i = 5+12+9+x+20+14+22 = 82+x \] Calculation of \(f_i x_i\):

\(x_i\)	\(f_i\)	\(f_i x_i\)
22	5	110
23	12	276
24	9	216
25	\(x\)	\(25x\)
26	20	520
27	14	378
28	22	616

\[ \sum f_i x_i = 2116 + 25x \] Substitute into mean formula: \[ 25.78=\frac{2116+25x}{82+x} \] \[ 25.78(82+x)=2116+25x \] \[ 2113.96+25.78x=2116+25x \] \[ 0.78x=2.04 \] \[ x\approx2.6 \] Frequency must be a whole number. \[ x=3 \] Answer: The number 25 appeared 3 times.

Question 18

Weights of 12 boys (kg): \[ 39, 51, 38, 34, 29, 31, 42, 40, 52, 44, 41, 35 \]

(a) Range

Maximum weight = 52 Minimum weight = 29 \[ \text{Range} = 52-29 = 23 \] Range = 23 kg

(b) Mean weight

\[ \text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}} \] Sum of weights: \[ 39+51+38+34+29+31+42+40+52+44+41+35 = 476 \] \[ \text{Mean}=\frac{476}{12}=39.67 \] Mean weight ≈ 39.67 kg

(c) Weight of the fattest boy

Largest value in the data: \[ 52 \text{ kg} \] Answer: 52 kg

(d) Boys having weight more than mean

Mean ≈ 39.67 Weights greater than mean: \[ 51,\ 42,\ 40,\ 52,\ 44,\ 41 \] Number of such boys: \[ 6 \] Answer: 6 boys