The numbers of cobs produced by 360 maize plants are shown in the following table. No. cobs 0 1 2 3 4 No. plants f 11 75 185 81 8 Calculate the mean and the standard deviation.

Given: No. cobs x 0 1 2 3 4 No. plants f 11 75 185 81 8 We know that, mean x ¯ = ∑ x f ∑ f Now, mean = 720 300 = 2 Now, Mean of squares = ∑ x 2 f ∑ f = 1672 360 = 4 . 64 We know that, SD = Mean of squares - Mean 2 Therefore, Standard deviation SD = 1672 300 - 720 300 2 = 0 . 803

The numbers of cobs produced by 360 maize plants are shown in the following table. No. cobs 0 1 2 3 4 No. plants f 11 75 185 81 8 Find the interquartile range and give an example of what it tells us about this dataset that the standard deviation does not tell us.

Given: No. cobs x 0 1 2 3 4 No. plants f 11 75 185 81 8 cumulative frequency 11 86 271 352 360 Now, lower quartile Q 1 is the median of the lower 50 % of the data. Here, there are 360 data values. If N is even, then the formula to find lower quartile is Q 1 = Average of N 4 th and N 4 + 1 th data values, where N is number of data values. ⇒ Lower quartile is average of 360 4 th and 360 4 + 1 th data values = Average of 90 th and 91 st data values. Here, the 90 th and 91 st data value is near to the cumulative frequency 271 , thus its corresponding data value is 2 . So, Q 1 = 2 + 2 2 = 2 Hence, the lower quartile of the given data is 2 . Now, upper quartile Q 3 is the median of the upper 50 % of the data. Here, there are 360 data values. If N is even, then the formula to find upper quartile is Q 3 = Average of 3 N 4 th and 3 N 4 + 1 th data values, where N is number of data values. ⇒ Upper quartile is average of 3 × 360 4 th and 3 × 360 4 + 1 th data values = Average of 270 th and 271 th data values. Here, the 270 th , 271 th data value is near the cumulative frequency 271 , so its corresponding data value is 2 ⇒ Q 3 = 2 + 2 2 = 2 Hence, the upper quartile of the given data is 2 . Here, Q 1 = Q 3 = 2 , so IQ R = Q 3 - Q 1 = 0 Thus, it tells us that the middle half ( 50 % ) of the values are identical, i.e., all equal to 2 .

The times spent, in minutes, by 30 girls and by 40 boys on an assignment are detailed in the following table. Time spent (min) 20 – 30 30 – 40 40 – 50 60 – 80 No. girls f 6 14 7 3 No. boys f 15 11 7 7 For the boys and for the girls, calculate estimates of the mean and standard deviation.

Given: Time spent in minutes 20 – 30 30 – 40 40 – 50 60 – 80 No. girls f 1 6 14 7 3 No. boys f 2 15 11 7 7 Class mid-values of time spend in minutes are 25 , 35 , 50 and 70 . Let the class mid- values be x . We know that, mean = ∑ x f 1 ∑ f 1 Now, girls mean = 1200 30 = 40 Now, girls Mean of squares = ∑ x 2 f 1 ∑ f 1 = 53100 30 = 1770 We know that, SD = Mean of squares - Mean 2 Therefore, girls Standard deviation SD = 53100 30 - 1200 30 2 = 13 . 0 minutes. Now, boys mean = ∑ x f 2 ∑ f 2 = 1600 40 = 40 Now, boys mean of squares = ∑ x 2 f 2 ∑ f 2 = 74660 40 = 1866 . 5 We know that, SD = Mean of squares - Mean 2 Therefore, boys Standard deviation SD = 74650 40 - 1600 40 2 = 16 . 3 minutes.

The times spent, in minutes, by 30 girls and by 40 boys on an assignment are detailed in the following table. Time spent (min) 20 – 30 – 40 – 60 – 80 No. girls f 6 14 7 3 No. boys f 15 11 7 7 It is required to make a comparison between the times spent by the two groups. What do the means tell us about the times spent?

Given: Time spent (min) 20 – 30 30 – 40 40 – 50 60 – 80 No. girls f 1 6 14 7 3 No. boys f 2 15 11 7 7 Class mid-values are 25 , 35 , 50 and 70 . Let the class mid- values be x . We know that, mean = ∑ x f 1 ∑ f 1 Now, girls mean = 1200 30 = 40 Now, girls Mean of squares = ∑ x 2 f 1 ∑ f 1 = 53100 30 = 1770 We know that, SD = Mean of squares - Mean 2 Therefore, girls Standard deviation SD = 53100 30 - 1200 30 2 = 13 . 0 minutes. Now, boys mean = ∑ x f 2 ∑ f 2 = 1600 40 = 40 Now, boys mean of squares = ∑ x 2 f 2 ∑ f 2 = 74660 40 = 1866 . 5 We know that, SD = Mean of squares - Mean 2 Therefore, boys Standard deviation SD = 74650 40 - 1600 40 2 = 16 . 3 minutes. Thus, on average, the times spent were very similar, as both means are estimated to be 40 minutes.

The times spent, in minutes, by 30 girls and by 40 boys on an assignment are detailed in the following table. Time spent (min) 20 – 30 – 40 – 60 – 80 No. girls f 6 14 7 3 No. boys f 15 11 7 7 It is required to make a comparison between the times spent by the two groups. Use the standard deviations to compare the times spent by the two groups.

Given: Time spent in minutes 20 – 30 30 – 40 40 – 50 60 – 80 No. girls f 1 6 14 7 3 No. boys f 2 15 11 7 7 Class mid-values of time spend in minutes are 25 , 35 , 50 and 70 . Let the class mid- values be x . We know that, mean = ∑ x f 1 ∑ f 1 Now, girls mean = 1200 30 = 40 Now, girls Mean of squares = ∑ x 2 f 1 ∑ f 1 = 53100 30 = 1770 We know that, SD = Mean of squares - Mean 2 Therefore, girls Standard deviation SD = 53100 30 - 1200 30 2 = 13 . 0 minutes. Now, boys mean = ∑ x f 2 ∑ f 2 = 1600 40 = 40 Now, boys mean of squares = ∑ x 2 f 2 ∑ f 2 = 74660 40 = 1866 . 5 We know that, SD = Mean of squares - Mean 2 Therefore, boys Standard deviation SD = 74650 40 - 1600 40 2 = 16 . 3 minutes. Thus, time spent by boys are more varied than time spent by girls.

The lengths, correct to the nearest centimetre, of 50 rods are given in the following table. Length cm 15 – 17 18 – 24 25 – 29 30 – 37 No. rods f 13 18 11 8 Calculate an estimate of the standard deviation of the lengths.

Given: First convert discrete class into continuous classes by subtracting 0 . 5 from the lower limit and adding 0 . 5 to the upper limit. length cm 14 . 5 - 17 . 5 17 . 5 - 24 . 5 24 . 5 - 29 . 5 29 . 5 - 37 . 5 No. rods f 13 18 11 8 Class mid-values are 16 , 21 , 27 and 33 . 5 . Let the class mid- values be x . We know that, mean x ¯ = ∑ x f ∑ f Now, mean = 1151 50 = 23 . 02 Now, Mean of squares = ∑ x 2 f ∑ f = 28263 50 = 565 . 26 We know that, SD x = Mean of squares - Mean 2 Therefore, Standard deviation SD x = 25263 50 - 1151 50 2 = 5 . 94 cm .

For the dataset denoted by x in the following table, k is a constant. x 15 16 17 18 19 20 f 2 k k + 5 k – 3 10 8 3 Find the value of k and calculate the variance of x , given that x ¯ = 17 .

Given: x 15 16 17 18 19 20 f 2 k k + 5 k – 3 10 8 3 We know that, mean x ¯ = ∑ x f ∑ f Now, 30 k + 16 k + 80 + 17 k - 51 + 180 + 152 + 60 4 k + 23 = 17 ⇒ 63 k + 421 4 k + 23 = 17 ⇒ 63 k + 421 = 68 k + 391 ⇒ k = 6 Now, mean of squares = ∑ x 2 f ∑ f = 13711 47 = 291 . 72 We know that, Variance = Mean of squares - Mean 2 Therefore, variance Var ( x ) = 13711 47 - 17 2 = 2 . 72 .

The following table illustrates the heights, in centimetres, of 150 children. Height cm No. children f 140 up to 144 a 144 up to 150 b 150 up to 160 69 160 up to 165 28 Given that a calculated estimate of the mean height is exactly 153 . 14 cm , show that 142 a + 147 b = 7726 , and evaluate a and b .

Given: Height cm No. children f 140 up to 144 a 144 up to 150 b 150 up to 160 69 160 up to 165 28 Class mid-values are 142 , 147 , 155 and 162 . 5 . Let the class mid- values be x . We know that, mean x ¯ = ∑ x f ∑ f Now, 142 a + 147 b + ( 155 × 69 ) + ( 162 . 5 × 28 ) 150 = 153 . 14 ⇒ 142 a + 147 b + 15245 = 22971 ⇒ 142 a + 147 b = 7726 .......(i) Hence showed. Here, total number of children = 150 ⇒ a + b + 69 + 28 = 150 ⇒ a + b = 53 ⇒ b = 53 - a .....(ii) Substitute equation (ii) in (i) ⇒ 142 a + 147 ( 53 - a ) = 7726 ⇒ a = 13 and b = 40 .

EASY

AS and A Level

IMPORTANT

Earn 100

The following table shows the number of pets owned by each of $35$ families.

No. pets $0$ $1$ $2$ $3$ $4$ $5$

No. families $(f)$ $6$ $12$ $9$ $4$ $3$ $1$

Find the mean and variance of the number of pets.

Important Questions on Measures of Variation

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 3 - Measures of Variation>EXERCISE 3B>Q 4

The numbers of cobs produced by $360$ maize plants are shown in the following table.

No. cobs	$0$	$1$	$2$	$3$	$4$
No. plants $(f)$	$11$	$75$	$185$	$81$	$8$

Calculate the mean and the standard deviation.

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 3 - Measures of Variation>EXERCISE 3B>Q 4

The numbers of cobs produced by $360$ maize plants are shown in the following table.

No. cobs	$0$	$1$	$2$	$3$	$4$
No. plants $(f)$	$11$	$75$	$185$	$81$	$8$

Find the interquartile range and give an example of what it tells us about this dataset that the standard deviation does not tell us.

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 3 - Measures of Variation>EXERCISE 3B>Q 5

The times spent, in minutes, by $30$ girls and by $40$ boys on an assignment are detailed in the following table.

Time spent (min)	$20 - 30$	$30 - 40$	$40 - 50$	$60 - 80$
No. girls $(f)$	$6$	$14$	$7$	$3$
No. boys $(f)$	$15$	$11$	$7$	$7$

For the boys and for the girls, calculate estimates of the mean and standard deviation.

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 3 - Measures of Variation>EXERCISE 3B>Q 5

The times spent, in minutes, by $30$ girls and by $40$ boys on an assignment are detailed in the following table.

Time spent (min)	$20 -$	$30 -$	$40 -$	$60 - 80$
No. girls $(f)$	$6$	$14$	$7$	$3$
No. boys $(f)$	$15$	$11$	$7$	$7$

It is required to make a comparison between the times spent by the two groups.

What do the means tell us about the times spent?

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 3 - Measures of Variation>EXERCISE 3B>Q 5

The times spent, in minutes, by $30$ girls and by $40$ boys on an assignment are detailed in the following table.

Time spent (min)	$20 -$	$30 -$	$40 -$	$60 - 80$
No. girls $(f)$	$6$	$14$	$7$	$3$
No. boys $(f)$	$15$	$11$	$7$	$7$

It is required to make a comparison between the times spent by the two groups.

Use the standard deviations to compare the times spent by the two groups.

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 3 - Measures of Variation>EXERCISE 3B>Q 6

The lengths, correct to the nearest centimetre, of $50$ rods are given in the following table.

Length $(cm)$	$15 - 17$	$18 - 24$	$25 - 29$	$30 - 37$
No. rods $(f)$	$13$	$18$	$11$	$8$

Calculate an estimate of the standard deviation of the lengths.

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 3 - Measures of Variation>EXERCISE 3B>Q 7

For the dataset denoted by $x$ in the following table, $k$ is a constant.

$x$	$15$	$16$	$17$	$18$	$19$	$20$
$f$	$2 k$	$k + 5$	$k - 3$	$10$	$8$	$3$

Find the value of $k$ and calculate the variance of $x,$ given that $\bar{x} = 17 .$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 3 - Measures of Variation>EXERCISE 3B>Q 8

The following table illustrates the heights, in centimetres, of $150$ children.

Height $(cm)$	No. children $(f)$
$140$ up to $144$	$a$
$144$ up to $150$	$b$
$150$ up to $160$	$69$
$160$ up to $165$	$28$

Given that a calculated estimate of the mean height is exactly $153.14 cm,$ show that $142 a + 147 b = 7726,$ and evaluate $a$ and $b .$

No. pets	$0$	$1$	$2$	$3$	$4$	$5$
No. families $(f)$	$6$	$12$	$9$	$4$	$3$	$1$

The following table shows the number of pets owned by each of 35 families. No. pets 0 1 2 3 4 5 No. families f 6 12 9 4 3 1 Find the mean and variance of the number of pets.

Important Questions on Measures of Variation

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The following table shows the number of pets owned by each of $35$ families.

No. pets $0$ $1$ $2$ $3$ $4$ $5$

No. families $(f)$ $6$ $12$ $9$ $4$ $3$ $1$

Find the mean and variance of the number of pets.