The box-and-whisker diagrams show the ages of people watching two different films at a cinema. Deduce which film is categorised as a family film. Give reasons for your answer.

Given a box-and-whisker plot for each data set using one number line for both plots: In a box plot, we draw a box from the first quartile to the third quartile. Box plot has the widest spread for the middle 50 % of the data (the data between the first and third quartiles). The first quartile marks one end of the box and the third quartile marks the other end of the box. Approximately the middle 50 % of the data fall inside the box. Compare the data between the first quartile and third quartile.The more the data is present between first and third quartile, the more is the data spread. General diagram of Box and whisker plot: For film A : Minimum value = 9 , maximum value = 65 , median = 38 years, upper quartile Q 3 = 49 years and lower quartile Q 1 = 39 years. For film B : Minimum value = 18 , maximum value = 46 , median = 32 years, upper quartile Q 3 = 41 years and lower quartile Q 1 = 25 years. Thus, the film A is categorised as a family film because both young and old people attended.

The table shows the masses (in kilograms) of Blackface sheep on two farms. Compare the masses of the Blackface sheep on the two farms.

Given data is Farm A 63 48 60 55 45 49 19 55 Farm B 54 71 68 57 62 70 54 49 Farm A 65 49 57 56 64 43 64 48 Farm B 66 68 72 50 56 49 70 64 Arranging the data in ascending order: Farm A 19 43 45 48 48 49 49 55 Farm A 55 56 57 60 63 64 64 65 Farm B 49 49 50 54 54 56 57 62 Farm B 64 66 68 68 70 70 71 72 A back-to-back stem-and-leaf plot is a method for comparing two data distributions by attaching two sets of 'leaves' to the same 'stem' in a stem-and-leaf plot. The leaf goes in the centre, with numbers running from least (at the top) to highest (at the bottom). Numbers in the right leaf run lowest to highest, with the lowest number placed closest to the stem. The data on the left side of the stem runs in the opposite direction (like a mirror) with the numbers running from lowest to highest starting at the stem. Here, the stem values are 1 , 2 , 3 , 4 , 5 , 6 , 7 . Median: Let the number of data values be N . Here there are 16 data values in both farm A and farm B . We know that, Median is average of N 2 th and N 2 + 1 th data value. Median = 8 th data value + 9 th data value 2 The median time for Farm A is = 55 + 55 2 = 55 . The median time for Farm B is 62 + 64 2 = 63 . Now, lower quartile Q 1 is the median of the lower 50 % of the data. Lower quartile is the average of N 4 th and N 4 + 1 th data values, where N is number of data values. ⇒ Lower quartile is average of 16 4 th and 16 4 + 1 th data values = Average of 4 th and 5 th data values. The lower quartile time for farm A is = 48 + 48 2 = 48 The lower quartile time for farm B is = 54 + 54 2 = 54 Now, upper quartile Q 3 is the median of the upper 50 % of the data. Upper quartile is average of 3 N 4 th and 3 N 4 + 1 th data values. ⇒ Upper quartile is average of 3 × 16 4 th and 3 × 16 4 + 1 th data values = Average of 12 th and 13 th data values. The upper quartile time for firm A is = 60 + 63 2 = 61 . 5 The upper quartile time for firm B is = 68 + 70 2 = 69 Interquartile range IQR = Upper quartile - Lower quartile . The interquartile range for Firm A is = 61 . 5 - 48 = 13 . 5 . The interquartile range for firm B is = 69 - 54 = 15 . Thus, farm B had a higher average and greater spread.

The table shows the masses (in kilograms) of Blackface sheep on two farms. The table shows the average adult bodyweight of Blackface sheep for different grazing conditions. Poor hill 45 - 50 kg Average/good hill 50 - 65 kg Upland 70 kg Deduce the grazing conditions available to the sheep on the two farms. Give reasons for your answer.

Given data is Farm A 63 48 60 55 45 49 19 55 Farm B 54 71 68 57 62 70 54 49 Farm A 65 49 57 56 64 43 64 48 Farm B 66 68 72 50 56 49 70 64 Arranging the data in ascending order: Farm A 19 43 45 48 48 49 49 55 Farm A 55 56 57 60 63 64 64 65 Farm B 49 49 50 54 54 56 57 62 Farm B 64 66 68 68 70 70 71 72 A back-to-back stem-and-leaf plot is a method for comparing two data distributions by attaching two sets of 'leaves' to the same 'stem' in a stem-and-leaf plot. The leaf goes in the centre, with numbers running from least (at the top) to highest (at the bottom). Numbers in the right leaf run lowest to highest, with the lowest number placed closest to the stem. The data on the left side of the stem runs in the opposite direction (like a mirror) with the numbers running from lowest to highest starting at the stem. Here, the stem values are 1 , 2 , 3 , 4 , 5 , 6 , 7 . Range of the data is difference between maximum data value and minimum data value. For firm A : Maximum value is 65 , minimum value is 19 . Thus, range for firm A = 65 - 19 = 46 . For firm B : Maximum value is 65 , minimum value is 19 . Thus, range for firm B = 49 - 72 = 23 . Therefore, Both the farms look like poor hill because the range of values fits that profile.

Rose Harrison and Clara Huizink Solutions for Exercise 22: Mixed practice

Author:Rose Harrison & Clara Huizink

Rose Harrison Mathematics Solutions for Exercise - Rose Harrison and Clara Huizink Solutions for Exercise 22: Mixed practice

Attempt the practice questions from Exercise 22: 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 22: Mixed practice with Hints & Solutions

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 5 - A Whole Range of Things (Working with Sets of data)>Mixed practice>Q 8

The box-and-whisker diagrams show the ages of people watching two different films at a cinema.

Question Image

Deduce which film is categorised as a family film. Give reasons for your answer.

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 5 - A Whole Range of Things (Working with Sets of data)>Mixed practice>Q 9

The profits made by a small business during the last $11$ years are as follows:

$$ 45000, $ 560000, $ 1000, $ 85000, $ 160000 $, $ 170000, $ 62000, $ 250000, $ 3100, $ 120000, $ 38000$

Find the interquartile range.(Answer without units)

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 5 - A Whole Range of Things (Working with Sets of data)>Mixed practice>Q 9

The profits made by a small business during the last $11$ years are as follows:
$$ 45000, $ 560000, $ 1000, $ 85000, $ 160000, $ 170000, $ 62000, $ 250000, $ 3100, $ 120000, $ 38000$

Determine if any of the data values can be considered outliers.(Answer without units)

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 5 - A Whole Range of Things (Working with Sets of data)>Mixed practice>Q 9

The profits made by a small business during the last $11$ years are as follows:
$$ 45000, $ 560000, $ 1000, $ 85000, $ 160000, $ 170000, $ 62000, $ 250000, $ 3100, $ 120000, $ 38000$

Find the range, Give a reason why any outliers have been included or excluded in your calculation.(outlier is $$ 560000$ )

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 5 - A Whole Range of Things (Working with Sets of data)>Mixed practice>Q 9

The profits made by a small business during the last $11$ years are as follows:
$$ 45000, $ 560000, $ 1000, $ 85000, $ 160000, $ 170000, $ 62000, $ 250000, $ 3100, $ 120000, $ 38000$

Find the measure of central tendency which is most appropriate in this case. Justify your choice.(outlier $$ 560000$ )

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 5 - A Whole Range of Things (Working with Sets of data)>Mixed practice>Q 10

The table shows the masses (in kilograms) of Blackface sheep on two farms.

$\begin{array}{l} Farm A & 63 & 48 & 60 & 55 & 45 & 49 & 19 & 55 \\ Farm B & 54 & 71 & 68 & 57 & 62 & 70 & 54 & 49 \\ Farm A & 65 & 49 & 57 & 56 & 64 & 43 & 64 & 48 \\ Farm B & 66 & 68 & 72 & 50 & 56 & 49 & 70 & 64 \end{array}$

Construct a back-to-back stem-and-leaf diagram to represent the data.

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 5 - A Whole Range of Things (Working with Sets of data)>Mixed practice>Q 10

The table shows the masses (in kilograms) of Blackface sheep on two farms.

Question Image

Compare the masses of the Blackface sheep on the two farms.

EASY

MYP:4-5

IMPORTANT

MYP Mathematics A concept based approach 4&5 Standard>Chapter 5 - A Whole Range of Things (Working with Sets of data)>Mixed practice>Q 10

The table shows the masses (in kilograms) of Blackface sheep on two farms.

Question Image

The table shows the average adult bodyweight of Blackface sheep for different grazing conditions.

$\begin{array}{l} Poor hill & 45 - 50 kg \\ Average/good hill & 50 - 65 kg \\ Upland & 70 kg \end{array}$

Deduce the grazing conditions available to the sheep on the two farms. Give reasons for your answer.

Rose Harrison and Clara Huizink Solutions for Exercise 22: Mixed practice

Rose Harrison Mathematics Solutions for Exercise - Rose Harrison and Clara Huizink Solutions for Exercise 22: Mixed practice

Questions from Rose Harrison and Clara Huizink Solutions for Exercise 22: Mixed practice with Hints & Solutions

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