The doctors at a medical center see 80 patients in total during morning surgery. The table gives information about the lengths of these appointments in minutes. Length, L (min) Frequency 0 < L ≤ 5 7 5 < L ≤ 7 15 7 < L ≤ 9 24 9 < L ≤ 13 19 13 < L ≤ 16 9 16 < L ≤ 20 6 Construct a frequency density histogram for the information.

Given data is length of appointment times taken by patients Class width of a class interval is difference between upper and lower boundaries of that class. Frequency density of class interval = Frequency class width So, the given table is Length, L (min) Frequency Class width Frequency density 0 < L ≤ 5 7 5 - 0 = 5 7 5 = 1 . 4 5 < L ≤ 7 15 7 - 5 = 2 15 2 = 7 . 5 7 < L ≤ 9 24 9 - 7 = 2 24 2 = 12 9 < L ≤ 13 19 4 19 4 = 4 . 75 13 < L ≤ 16 9 3 9 3 = 3 16 < L ≤ 20 6 4 6 4 = 1 . 5 Now frequency density histogram is constructed by plotting appointment length on x - axis and frequency density on y - axis. Draw the bars with no gaps between them. The frequency density histogram is as follows

The table shows the times taken by members of a cycling club to finish a 100 km charity ride. Time, t (min) Frequency 110 < t ≤ 130 6 130 < t ≤ 145 30 145 < t ≤ 160 45 160 < t ≤ 185 55 185 < t ≤ 195 28 195 < t ≤ 250 16 Construct a frequency density histogram for the information.

Given data is times taken by members of a club. Class width of a class interval is difference between upper and lower boundaries of that class. Frequency density of class interval = Frequency class width So, the given table is Time, t (min) Frequency Class width Frequency density 110 < t ≤ 130 6 130 - 110 = 20 6 20 = 0 . 3 130 < t ≤ 145 30 145 - 130 = 15 30 15 = 2 145 < t ≤ 160 45 160 - 145 = 15 45 15 = 3 160 < t ≤ 185 55 185 - 160 = 25 55 25 = 2 . 2 185 < t ≤ 195 28 195 - 185 = 10 28 10 = 2 . 8 195 < t ≤ 250 16 250 - 195 = 55 16 55 = 0 . 29 Now frequency density histogram is constructed by plotting time on x - axis and frequency density on y - axis. Draw the bars with no gaps between them. The frequency density histogram is as follows

The table shows the times taken by members of a cycling club to finish a 100 km charity ride. Time, t (min) Frequency 110 < t ≤ 130 6 130 < t ≤ 145 30 145 < t ≤ 160 45 160 < t ≤ 185 55 185 < t ≤ 195 28 195 < t ≤ 250 16 State the modal class.

Given data is times taken by members of a club. Class width of a class interval is difference between upper and lower boundaries of that class. Frequency density of class interval = Frequency class width So, the given table is Time, t (min) Frequency Class width Frequency density 110 < t ≤ 130 6 130 - 110 = 20 6 20 = 0 . 3 130 < t ≤ 145 30 145 - 130 = 15 30 15 = 2 145 < t ≤ 160 45 160 - 145 = 15 45 15 = 3 160 < t ≤ 185 55 185 - 160 = 25 55 25 = 2 . 2 185 < t ≤ 195 28 195 - 185 = 10 28 10 = 2 . 8 195 < t ≤ 250 16 250 - 195 = 55 16 55 = 0 . 29 The modal class of data with varying widths is the class interval with highest frequency density. So, 145 < x ≤ 160 is modal class as it has frequency density of value 3 .

The histogram represents the distances students threw the javelin on school sports day. Find the class interval that contains the median.

Given data is distances students threw javelin Class width of a class interval is difference between upper and lower boundaries of that class. Frequency density of class interval = Frequency class width . So, frequency of class is product of class width and frequency density. So, the given table is Age, x (years) Frequency density Class width Frequency Cumulative frequency 0 ≤ x < 20 3 20 - 0 = 20 20 × 3 = 60 60 20 ≤ x < 35 6 35 - 20 = 15 15 × 6 = 90 150 35 ≤ x < 45 10 45 - 35 = 10 10 × 10 = 100 250 45 ≤ x < 60 6 60 - 45 = 15 15 × 6 = 90 340 60 ≤ x < 65 2 65 - 60 = 5 5 × 2 = 10 350 Now, let n = sum of all frequencies n = 350 as n is even, so the position of median is the n 2 th term Therefore, the position of median is = 350 2 = 175 th term of frequency So, the frequency of 175 will be possible in 35 - 45 interval. So, the median will be in the interval 35 - 45 .

Rose Harrison and Clara Huizink Solutions for Exercise 16: Practice 5

Author:Rose Harrison & Clara Huizink

Rose Harrison Mathematics Solutions for Exercise - Rose Harrison and Clara Huizink Solutions for Exercise 16: Practice 5

Attempt the practice questions from Exercise 16: Practice 5 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 16: Practice 5 with Hints & Solutions

MEDIUM

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 11 - City Skylines (Histograms)>Practice 5>Q 3

The doctors at a medical center see $80$ patients in total during morning surgery. The table gives information about the lengths of these appointments in minutes.

Length, $L$ (min)	Frequency
$0 < L \leq 5$	$7$
$5 < L \leq 7$	$15$
$7 < L \leq 9$	$24$
$9 < L \leq 13$	$19$
$13 < L \leq 16$	$9$
$16 < L \leq 20$	$6$

Construct a frequency density histogram for the information.

MEDIUM

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 11 - City Skylines (Histograms)>Practice 5>Q 3

The doctors at a medical center see $80$ patients in total during morning surgery. The table gives information about the lengths of these appointments in minutes.

Length, $L$ (min)	Frequency
$0 < L \leq 5$	$7$
$5 < L \leq 7$	$15$
$7 < L \leq 9$	$24$
$9 < L \leq 13$	$19$
$13 < L \leq 16$	$9$
$16 < L \leq 20$	$6$

Calculate an estimate for the number of patients who saw their doctor for between $10$ and $15$ minutes.

Number of patients should be a integer without any decimal places, so round it off to zero decimal places.

MEDIUM

MYP:4-5

IMPORTANT

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

The table shows the times taken by members of a cycling club to finish a $100 km$ charity ride.

Time, $t$ (min)	Frequency
$110 < t \leq 130$	$6$
$130 < t \leq 145$	$30$
$145 < t \leq 160$	$45$
$160 < t \leq 185$	$55$
$185 < t \leq 195$	$28$
$195 < t \leq 250$	$16$

Construct a frequency density histogram for the information.

EASY

MYP:4-5

IMPORTANT

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

The table shows the times taken by members of a cycling club to finish a $100 km$ charity ride.

Time, $t$ (min)	Frequency
$110 < t \leq 130$	$6$
$130 < t \leq 145$	$30$
$145 < t \leq 160$	$45$
$160 < t \leq 185$	$55$
$185 < t \leq 195$	$28$
$195 < t \leq 250$	$16$

State the modal class.

MEDIUM

MYP:4-5

IMPORTANT

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

The table shows the times taken by members of a cycling club to finish a $100 km$ charity ride.

Time, $t$ (min)	Frequency
$110 < t \leq 130$	$6$
$130 < t \leq 145$	$30$
$145 < t \leq 160$	$45$
$160 < t \leq 185$	$55$
$185 < t \leq 195$	$28$
$195 < t \leq 250$	$16$

Calculate an estimate for the number of cyclists who took between $2 \frac{1}{2}$ and $3 \frac{1}{2}$ hours to complete the ride.

Given the answer by rounding off to zero decimal places.

EASY

MYP:4-5

IMPORTANT

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

The histogram represents the distances students threw the javelin on school sports day.

Question Image

Find the class interval that contains the median.

EASY

MYP:4-5

IMPORTANT

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

This histogram represents children's heights. Fifteen children measured between $110 cm$ and $125 cm$ .

Question Image

Frequency density at interval $125 - 130$ is $2.2$ and at $140 - 150$ is $2.4$ .

Find the number of children in the sample.

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 11 - City Skylines (Histograms)>Practice 5>Q 7

In a histogram, a bar of width $2 cm$ and height $5 cm$ represents a frequency of $50$ . A second bar is $4 cm$ wide and represents a frequency of $50$ . Calculate the height of this bar in $cm$ .

Rose Harrison and Clara Huizink Solutions for Exercise 16: Practice 5

Rose Harrison Mathematics Solutions for Exercise - Rose Harrison and Clara Huizink Solutions for Exercise 16: Practice 5

Questions from Rose Harrison and Clara Huizink Solutions for Exercise 16: Practice 5 with Hints & Solutions

Learn and Prepare for any exam you want !