The heights of 14 -year-old students from Sri Lanka and Peru are given in the frequency tables below. Heights of students in Sri Lanka (measured to the nearest cm) Frequency 120 - 129 3 130 - 139 11 140 - 149 14 150 - 159 8 160 - 169 6 Heights of students in Peru (measured to the nearest cm) Frequency 120 - 129 22 130 - 139 41 140 - 149 32 150 - 159 6 160 - 169 6 Explain why it is necessary to construct a relative frequency histogram to compare these data sets.

Given data shows the heights of students from Sri Lanka and Peru In order to make fair comparisons between data sets with different numbers of data points, we can use a relative frequency histogram. Here we have different number of data points in each group, so we need to use relative frequency histogram. Relative frequency is the proportion (or percentage) of the data set belonging in the class interval. For a data set with n members, a class interval with frequency f has relative frequency f n .

The heights of 14 -year-old students from Sri Lanka and Peru are given in the frequency tables below. Heights of students in Sri Lanka (measured to the nearest cm) Frequency 120 - 129 3 130 - 139 11 140 - 149 14 150 - 159 8 160 - 169 6 Heights of students in Peru (measured to the nearest cm) Frequency 120 - 129 22 130 - 139 41 140 - 149 32 150 - 159 6 160 - 169 6 Comment on the distributions of the heights of 14 -year-old students in Sri Lanka and Peru.

Given data shows the heights of students from Sri Lanka and Peru For a data set with n members or total frequency n , a class interval with frequency f has relative frequency f n . Now the relative frequency table of students from sri lanka whose total frequency is = 3 + 11 + 14 + 8 + 6 = 42 is given by Heights of students in Sri Lanka (measured to the nearest cm) Frequency f Relative frequency (Rounded off to two decimal places) 120 - 129 3 3 42 = 0 . 07 130 - 139 11 11 42 = 0 . 26 140 - 149 14 14 42 = 0 . 33 150 - 159 8 8 42 = 0 . 19 160 - 169 6 6 42 = 0 . 14 Now the relative frequency table for students in peru whose total frequency is = 22 + 41 + 32 + 6 + 6 = 107 is given by Heights of students in Peru (measured to the nearest cm) Frequency Relative frequency (Rounded off to two decimal places) 120 - 129 22 22 107 = 0 . 21 130 - 139 41 41 107 = 0 . 38 140 - 149 32 32 107 = 0 . 30 150 - 159 6 6 107 = 0 . 06 160 - 169 6 6 107 = 0 . 06 Now plot the Heights on x - axis and relative frequency on y - axis to construct the Relative frequency histogram of data of students in sri lanka. Now plot the Heights on x - axis and relative frequency on y - axis to construct the Relative frequency histogram of data of students in peru. The histogram for students in sri lanka is fairly symmetrical and it is a unimodal distribution as we have only one peak. The histogram for students in peru is a unimodal distribution as we have only one peak and it is not so symmetrical.

In New Zealand, 90 mothers were asked how old they were when they had their first child. The frequency table gives the results. Age, x (years) Frequency 15 < x ≤ 20 6 20 < x ≤ 22 14 22 < x ≤ 25 19 25 < x ≤ 30 21 30 < x ≤ 40 30 Construct the histogram.

Class width of a class interval is difference between upper and lower boundaries of that class. for class interval 15 < x ≤ 20 , class width = 20 - 15 = 5 , for class interval 20 < x ≤ 22 , class width = 22 - 20 = 2 , Here, we have varying class widths in this data, so frequency density histogram would give the best representation for this data. Frequency density of class interval = Frequency class width So, the given table is Age, x (years) Frequency Class width Frequency density 15 < x ≤ 20 6 5 6 5 = 1 . 2 20 < x ≤ 22 14 2 14 2 = 7 22 < x ≤ 25 19 3 19 3 = 6 . 3 25 < x ≤ 30 21 5 21 5 = 4 . 2 30 < x ≤ 40 30 10 30 10 = 3 Now frequency density histogram is constructed by plotting age on x - axis and frequency density on y - axis. Draw the bars with no gaps between them. The frequency density histogram is as follows

In New Zealand, 90 mothers were asked how old they were when they had their first child. The frequency table gives the results. Age, x (years) Frequency 15 < x ≤ 20 6 20 < x ≤ 22 14 22 < x ≤ 25 19 25 < x ≤ 30 21 30 < x ≤ 40 30 Calculate an estimate for the mean and determine the modal class.

Class width of a class interval is difference between upper and lower boundaries of that class. for class interval 15 < x ≤ 20 , class width = 20 - 15 = 5 , for class interval 20 < x ≤ 22 , class width = 22 - 20 = 2 , Here, we have varying class widths in this data, so frequency density histogram would give the best representation for this data. Frequency density of class interval = Frequency class width For finding estimated mean age for this grouped data we make a column of midpoint of each class interval. So, the given table is Age, x (years) Frequency f Class width Frequency density Midpoint of interval y 15 < x ≤ 20 6 5 6 5 = 1 . 2 15 + 20 2 = 17 . 5 20 < x ≤ 22 14 2 14 2 = 7 22 + 20 2 = 21 22 < x ≤ 25 19 3 19 3 = 6 . 3 23 . 5 25 < x ≤ 30 21 5 21 5 = 4 . 2 27 . 5 30 < x ≤ 40 30 10 30 10 = 3 35 Now, the mean age is = ∑ f y ∑ f = 6 × 17 . 5 + 21 × 14 + 23 . 5 × 19 + 27 . 5 × 21 + 35 × 30 6 + 14 + 19 + 21 + 30 = 2473 90 = 27 . 5 (up to one decimal place) So, the estimated mean age is 27 . 5 years. The modal class of data with varying widths is the class interval with highest frequency density. So, 20 < x ≤ 22 is modal class as it has frequency density of value 7 .

The histogram shows the waiting times for some customers at a supermarket. Frequency density value at interval 0 - 40 is 0 . 2 , 90 - 120 is 0 . 7 , 120 - 180 is 0 . 3 , 180 - 240 is 0 . 1 State the modal class.

Given data is waiting time for some customers. So, from graph Waiting time, t (sec) Frequency density 0 - 40 0 . 2 40 - 60 0 . 5 60 - 70 1 . 5 70 - 90 1 90 - 120 0 . 7 120 - 180 0 . 3 180 - 240 0 . 1 The modal class of data with varying widths is the class interval with highest frequency density. So, 60 - 70 is modal class as it has frequency density of value 1 . 5 .

Rose Harrison and Clara Huizink Solutions for Exercise 17: Mixed practice

Author:Rose Harrison & Clara Huizink

Rose Harrison Mathematics Solutions for Exercise - Rose Harrison and Clara Huizink Solutions for Exercise 17: Mixed practice

Attempt the practice questions from Exercise 17: Mixed practice with hints and solutions to strengthen your understanding. MYP Mathematics A concept based approach 4&5 Standard solutions are prepared by Experienced Embibe Experts.

Questions from Rose Harrison and Clara Huizink Solutions for Exercise 17: Mixed practice with Hints & Solutions

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 11 - City Skylines (Histograms)>Mixed practice>Q 6

The heights of $14$ -year-old students from Sri Lanka and Peru are given in the frequency tables below.

Heights of students in Sri Lanka (measured to the nearest cm)	Frequency
$120 - 129$	$3$
$130 - 139$	$11$
$140 - 149$	$14$
$150 - 159$	$8$
$160 - 169$	$6$

Heights of students in Peru (measured to the nearest cm)	Frequency
$120 - 129$	$22$
$130 - 139$	$41$
$140 - 149$	$32$
$150 - 159$	$6$
$160 - 169$	$6$

Explain why it is necessary to construct a relative frequency histogram to compare these data sets.

MEDIUM

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 11 - City Skylines (Histograms)>Mixed practice>Q 6

The heights of $14$ -year-old students from Sri Lanka and Peru are given in the frequency tables below.

Heights of students in Sri Lanka (measured to the nearest cm)	Frequency
$120 - 129$	$3$
$130 - 139$	$11$
$140 - 149$	$14$
$150 - 159$	$8$
$160 - 169$	$6$

Heights of students in Peru (measured to the nearest cm)	Frequency
$120 - 129$	$22$
$130 - 139$	$41$
$140 - 149$	$32$
$150 - 159$	$6$
$160 - 169$	$6$

Comment on the distributions of the heights of $14$ -year-old students in Sri Lanka and Peru.

MEDIUM

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 11 - City Skylines (Histograms)>Mixed practice>Q 8

In New Zealand, $90$ mothers were asked how old they were when they had their first child. The frequency table gives the results.

Age, $x$ (years)	Frequency
$15 < x \leq 20$	$6$
$20 < x \leq 22$	$14$
$22 < x \leq 25$	$19$
$25 < x \leq 30$	$21$
$30 < x \leq 40$	$30$

Construct the histogram.

MEDIUM

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 11 - City Skylines (Histograms)>Mixed practice>Q 8

In New Zealand, $90$ mothers were asked how old they were when they had their first child. The frequency table gives the results.

Age, $x$ (years)	Frequency
$15 < x \leq 20$	$6$
$20 < x \leq 22$	$14$
$22 < x \leq 25$	$19$
$25 < x \leq 30$	$21$
$30 < x \leq 40$	$30$

Calculate an estimate for the mean and determine the modal class.

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 11 - City Skylines (Histograms)>Mixed practice>Q 10

The histogram shows the waiting times for some customers at a supermarket.

Question Image

Frequency density value at interval $0 - 40$ is $0.2$ , $90 - 120$ is $0.7$ , $120 - 180$ is $0.3$ , $180 - 240$ is $0.1$

State the modal class.

MEDIUM

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 11 - City Skylines (Histograms)>Mixed practice>Q 10

The histogram shows the waiting times for some customers at a supermarket.

Question Image

Frequency density value at interval $0 - 40$ is $0.2$ , $90 - 120$ is $0.7$ , $120 - 180$ is $0.3$ , $180 - 240$ is $0.1$

Calculate an estimate for the mean waiting time in seconds rounded off to two decimal places.

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 11 - City Skylines (Histograms)>Mixed practice>Q 10

The histogram shows the waiting times for some customers at a supermarket.

Question Image

Frequency density value at interval $0 - 40$ is $0.2$ , $90 - 120$ is $0.7$ , $120 - 180$ is $0.3$ , $180 - 240$ is $0.1$

Find the total number of customers.

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 11 - City Skylines (Histograms)>Mixed practice>Q 10

The histogram shows the waiting times for some customers at a supermarket.

Question Image

Frequency density value at interval $0 - 40$ is $0.2$ , $90 - 120$ is $0.7$ , $120 - 180$ is $0.3$ , $180 - 240$ is $0.1$

Find an estimate for the number of customers who had to wait between $60$ and $90$ seconds, and also find who had to wait longer than $2$ minutes.

Rose Harrison and Clara Huizink Solutions for Exercise 17: Mixed practice

Rose Harrison Mathematics Solutions for Exercise - Rose Harrison and Clara Huizink Solutions for Exercise 17: Mixed practice

Questions from Rose Harrison and Clara Huizink Solutions for Exercise 17: Mixed practice with Hints & Solutions

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