Solutions: Statistics - Test 2

Q1. (a) 3 Median = Mode + 2 Mean

This is the empirical relationship between Mean, Median, and Mode.

Q2. (a) 9

The number 9 appears most frequently (3 times) in the data set.

Q3. (b) frequency of class preceding modal class

$f_1$ is modal class frequency, $f_0$ is preceding, $f_2$ is succeeding.

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

Mode corresponds to the highest frequency, which is represented by the highest bar in a histogram.

Q5. Mean = 15, Median = 18.

Mode = 3 Median - 2 Mean
Mode = $3(18) - 2(15) = 54 - 30 = 24$.

Q6. Max frequency is 18. Modal class is 40-60.

$l=40, f_1=18, f_0=6, f_2=10, h=20$.
Mode = $40 + \frac{18-6}{2(18)-6-10} \times 20 = 40 + \frac{12}{36-16} \times 20$
$= 40 + \frac{12}{20} \times 20 = 40 + 12 = 52$.

Q7. Convert cumulative frequency to simple frequency.

0-10: 3
10-20: $12-3=9$
20-30: $27-12=15$
30-40: $57-27=30$ (Max Frequency)
40-50: $75-57=18$
Modal Class is 30-40.

Q8. Max frequency is 45. Modal class is 30-40.

$l=30, f_1=45, f_0=35, f_2=25, h=10$.
Mode = $30 + \frac{45-35}{2(45)-35-25} \times 10 = 30 + \frac{10}{90-60} \times 10$
$= 30 + \frac{10}{30} \times 10 = 30 + \frac{10}{3} = 30 + 3.33 = 33.33$.

Q9. Mode = 55. Modal class is 45-60.

$l=45, f_1=15, f_0=x, f_2=10, h=15$.
$55 = 45 + \frac{15-x}{2(15)-x-10} \times 15$
$10 = \frac{15-x}{30-x-10} \times 15 \Rightarrow 10 = \frac{15-x}{20-x} \times 15$
Divide by 5: $2 = \frac{15-x}{20-x} \times 3 \Rightarrow 2(20-x) = 3(15-x)$
$40 - 2x = 45 - 3x \Rightarrow 3x - 2x = 45 - 40 \Rightarrow x = 5$.

Q10. Max frequency is 61. Modal class is 60-80.

$l=60, f_1=61, f_0=52, f_2=38, h=20$.
Mode = $60 + \frac{61-52}{2(61)-52-38} \times 20$
$= 60 + \frac{9}{122-90} \times 20 = 60 + \frac{9}{32} \times 20$
$= 60 + \frac{9 \times 5}{8} = 60 + \frac{45}{8} = 60 + 5.625 = 65.625$ hours.

(i) Max frequency is 18. Modal Class: 145-150.
Total $N=51$. $N/2 = 25.5$. Cumulative freqs: 4, 11, 29... 29 > 25.5. Median Class: 145-150.

(ii) Mode Calculation:
$l=145, f_1=18, f_0=7, f_2=11, h=5$.
Mode = $145 + \frac{18-7}{36-7-11} \times 5 = 145 + \frac{11}{18} \times 5$
$= 145 + \frac{55}{18} = 145 + 3.05 = 148.05$ cm.

(iii) Mean = 149.02 cm. Mode = 148.05 cm.
Mean > Mode. The average height is slightly higher than the most common height.