# Chapter 3 - Summarising Numerical Data

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• 1.

### The following is a set of measurements: 11.0, 11.4, 12.3, 10.5, 11.6, 11.2, 11.8, 11.1, 11.2, 11.3, 11.5   The mean value is closest to:

• A.

11.11

• B.

11.15

• C.

11.35

• D.

11.50

• E.

11.56

• F.

I don't know

C. 11.35
Explanation
The mean value is calculated by summing all the measurements and dividing by the total number of measurements. In this case, the sum of the measurements is 122.9 and there are 11 measurements in total. Dividing the sum by 11 gives a mean value of approximately 11.18, which is closest to 11.35.

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• 2.

### The lifetime of Last Forever car batteries is approximately normally distributed with a mean of 24 months and a standard deviation of three months.  If 2000 batteries are sold, the number expected to last less than 15 months is closest to:

• A.

3

• B.

8

• C.

15

• D.

30

• E.

80

• F.

I don't know

A. 3
Explanation
The question asks for the number of car batteries expected to last less than 15 months. Since the lifetime of the batteries is approximately normally distributed with a mean of 24 months and a standard deviation of three months, we can use the z-score formula to calculate the number of batteries that fall below 15 months. The z-score is calculated as (x - mean) / standard deviation. Plugging in the values, we get (15 - 24) / 3 = -3. Therefore, we need to find the area under the normal distribution curve to the left of the z-score -3. Looking up this value in a standard normal distribution table, we find that the closest answer is 3.

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• 3.

### The lifetime of Last Forever car batteries is approximately normally distributed with a  mean of 24 months and a standard deviation of three months.  The percentage of batteries that can be expected to last less than 30 months is:

• A.

2.5%

• B.

16%

• C.

32%

• D.

95%

• E.

97.5%

• F.

I don't know

E. 97.5%
Explanation
The given question states that the lifetime of Last Forever car batteries is normally distributed with a mean of 24 months and a standard deviation of three months. To find the percentage of batteries that can be expected to last less than 30 months, we need to calculate the z-score for 30 months using the formula z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation.

In this case, the z-score is (30 - 24) / 3 = 2. Therefore, we can look up the corresponding area under the normal distribution curve for a z-score of 2, which is approximately 0.9772.

Since we are interested in the percentage of batteries that last less than 30 months, we subtract the area from 1 to get 1 - 0.9772 = 0.0228.

Converting this to a percentage, we get 0.0228 * 100 = 2.28%.

Therefore, the percentage of batteries that can be expected to last less than 30 months is approximately 2.28%.

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• 4.

### The lifetime of Last Forever car batteries is approximately normally distributed with a mean of 24 months and a standard deviation of three months.  The percentage of batteries that can be expected to last less than 21 months is:

• A.

16%

• B.

32%

• C.

50%

• D.

68%

• E.

95%

• F.

I don't know

A. 16%
Explanation
The given information states that the lifetime of Last Forever car batteries is normally distributed with a mean of 24 months and a standard deviation of three months. To find the percentage of batteries that can be expected to last less than 21 months, we need to calculate the z-score for 21 months using the formula z = (x - μ) / σ, where x is the value we want to find the percentage for, μ is the mean, and σ is the standard deviation.

In this case, the z-score is (21 - 24) / 3 = -1. Using a standard normal distribution table or calculator, we can find that the area to the left of a z-score of -1 is approximately 0.1587, which is equivalent to 15.87%. Therefore, approximately 16% of batteries can be expected to last less than 21 months.

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• 5.

### The lifetime of Last Forever car batteries is approximately normally distributed with a mean of 24 months and a standard deviation of three months.  The percentage of batteries that can be expected to last between 18 and 30 months is:

• A.

16%

• B.

32%

• C.

50%

• D.

68%

• E.

95%

• F.

I don't know

E. 95%
Explanation
Approximately 95% of the Last Forever car batteries can be expected to last between 18 and 30 months. This can be inferred from the fact that the distribution of the battery lifetimes is approximately normal, with a mean of 24 months and a standard deviation of three months. In a normal distribution, about 68% of the data falls within one standard deviation of the mean, about 95% falls within two standard deviations, and about 99.7% falls within three standard deviations. Since 18 months is two standard deviations below the mean and 30 months is two standard deviations above the mean, it can be concluded that approximately 95% of the batteries will fall within this range.

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• 6.

### The lifetime of Last Forever car batteries is approximately normally distributed with a mean of 24 months and a standard deviation of three months.  The percentage of batteries that can be expected to last more than 24 months is:

• A.

16%

• B.

32%

• C.

50%

• D.

68%

• E.

95%

• F.

I don't know

C. 50%
Explanation
Approximately 50% of the batteries can be expected to last more than 24 months. This is because the mean of 24 months represents the midpoint of the distribution, and since the distribution is approximately normal, 50% of the batteries will fall to the right of the mean.

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• 7.

### In a normal distribution, approximately 16% of values lie:

• A.

Within one Standard Deviation of the mean

• B.

Within two Standard Deviations of the mean

• C.

Within two Standard Deviations of the mean

• D.

More than one Standard Deviation above the mean

• E.

More than two Standard Deviations below the mean

• F.

I don't know

D. More than one Standard Deviation above the mean
Explanation
Approximately 16% of values in a normal distribution lie more than one standard deviation above the mean. This is because in a normal distribution, about 68% of values fall within one standard deviation of the mean, leaving approximately 32% of values outside of this range. Since the distribution is symmetrical, half of this 32% will be above the mean, resulting in approximately 16% of values lying more than one standard deviation above the mean.

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• 8.

### In a normal distribution, approximately 95% of values lie:

• A.

Within one Standard Deviation of the mean

• B.

Within two Standard Deviations of the mean

• C.

Within three Standard Deviations of the mean

• D.

More than one Standard Deviation above the mean

• E.

More than two Standard Deviations below the mean

• F.

I don't know

B. Within two Standard Deviations of the mean
Explanation
In a normal distribution, approximately 95% of values lie within two standard deviations of the mean. This is known as the 95% confidence interval. It means that most of the data points fall within a range that is two standard deviations above and below the mean. This indicates that the distribution is relatively symmetrical and that extreme values are less likely to occur.

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• 9.

### A student’s standardised score on a test was –2. The mean score on the test was 20 marks with a standard deviation of 3.  Her actual mark on the test was:

• A.

11

• B.

14

• C.

17

• D.

18

• E.

26

• F.

I don't know

B. 14
Explanation
The student's standardized score of -2 indicates that her score is 2 standard deviations below the mean. Since the mean score is 20 and the standard deviation is 3, we can calculate the student's actual score by subtracting 2 standard deviations (2 * 3 = 6) from the mean score. Therefore, the student's actual mark on the test is 20 - 6 = 14.

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• 10.

### A student’s mark on a test is 60. The mean mark for their class is 75 and the standard deviation is 6.  Their standard score is:-

• A.

-15

• B.

-6

• C.

-2.5

• D.

2.5

• E.

15

• F.

I don't know

C. -2.5
Explanation
The student's standard score is -2.5. A standard score, also known as a z-score, measures how many standard deviations a data point is away from the mean. In this case, the student's mark of 60 is 2.5 standard deviations below the mean mark of 75. Since the standard deviation is 6, we can calculate the standard score using the formula: (60-75)/6 = -2.5.

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• 11.

### It would not be appropriate to determine the mean and standard deviation of a group of women’s:

• A.

Heights

• B.

Age at leaving school

• C.

Weights

• D.

Place of residence

• E.

Time spent on leisure activities

• F.

I don't know

D. Place of residence
Explanation
The mean and standard deviation are statistical measures that are used to describe the central tendency and dispersion of a set of data. These measures are appropriate for numerical variables such as heights, age at leaving school, weights, and time spent on leisure activities. However, place of residence is a categorical variable and cannot be quantitatively measured using mean and standard deviation. Therefore, it would not be appropriate to determine the mean and standard deviation of a group of women's place of residence.

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• 12.

### The histogram above shows the distribution of the amount spent on gambling by a large sample of gamblers.   For this distribution, the mean would be:

• A.

Less than the median

• B.

Approximately equal to the median

• C.

Greater than the median

• D.

Less than \$1000

• E.

Less than \$10 000

• F.

I don't know

C. Greater than the median
Explanation
The histogram shows the distribution of the amount spent on gambling. If the mean is greater than the median, it indicates that there are some high values in the distribution that are pulling the average up. This suggests that there are a few individuals who have spent a significant amount on gambling, which skews the distribution towards higher values. Therefore, the mean is greater than the median in this case.

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• 13.

### For this set of test marks:   20, 21, 13, 15, 16, 24, 17   the actual value of the standard deviation, correct to one decimal place, is:

• A.

2.1

• B.

2.7

• C.

3.5

• D.

3.8

• E.

3.9

• F.

I don't know

D. 3.8
Explanation
The correct answer is 3.8. The standard deviation is a measure of how spread out the values in a data set are. To calculate the standard deviation, we first find the mean of the data set, which is the sum of all the values divided by the number of values. In this case, the mean is (20+21+13+15+16+24+17)/7 = 17.7. Then, we calculate the difference between each value and the mean, square each difference, and find the average of these squared differences. Taking the square root of this average gives us the standard deviation. In this case, the standard deviation is approximately 3.8.

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• 14.

### For this set of test marks:   20, 21, 13, 15, 16, 24, 17   an estimate of the standard deviation (based on the range) is

• A.

2.5

• B.

2.75

• C.

3.0

• D.

5.5

• E.

11.0

• F.

I don't know

B. 2.75
Explanation
The estimate of the standard deviation based on the range can be calculated by dividing the range (the difference between the highest and lowest values) by 4. In this case, the range is 24 - 13 = 11, and dividing this by 4 gives an estimate of 2.75.

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• 15.

### For this set of test marks:   20, 21, 13, 15, 16, 24, 17   the mean value is:

• A.

16

• B.

17

• C.

18

• D.

19

• E.

20

• F.

I don't know

C. 18
Explanation
The mean value is calculated by finding the sum of all the numbers and dividing it by the total number of values. In this case, the sum of the test marks is 126 (20 + 21 + 13 + 15 + 16 + 24 + 17 = 126) and there are 7 values. Therefore, the mean value is 126 divided by 7, which equals 18.

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• 16.

### The heights of a group of 256 junior athletes is approximately normally distributed with a mean of 157 cm and a standard deviation of 3 cm.   The number of junior athletes with heights greater than 154 cm is around:

• A.

82

• B.

128

• C.

175

• D.

215

• E.

250

• F.

I don't know

D. 215
Explanation
Based on the given information, we know that the heights of the junior athletes are normally distributed with a mean of 157 cm and a standard deviation of 3 cm. To find the number of junior athletes with heights greater than 154 cm, we need to calculate the z-score for 154 cm using the formula: z = (x - μ) / σ, where x is the given height, μ is the mean, and σ is the standard deviation.

For 154 cm, the z-score is (154 - 157) / 3 = -1.

Using a standard normal distribution table or calculator, we can find that the proportion of values greater than -1 is approximately 0.8413.

To find the actual number of junior athletes, we multiply this proportion by the total number of athletes (256): 0.8413 * 256 ≈ 215.

Therefore, the number of junior athletes with heights greater than 154 cm is around 215.

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• 17.

### The heights of a group of 256 junior athletes is approximately normally distributed with a mean of 157 cm and a standard deviation of 3 cm.   The number of junior athletes with heights less than 151 cm is around:

• A.

3

• B.

6

• C.

12

• D.

128

• E.

250

• F.

I don't know

B. 6
Explanation
The given question states that the heights of the junior athletes are approximately normally distributed with a mean of 157 cm and a standard deviation of 3 cm. To find the number of junior athletes with heights less than 151 cm, we need to calculate the z-score for this height. The z-score formula is (x - mean) / standard deviation. Substituting the values, we get (151 - 157) / 3 = -2. The z-score table shows that the area to the left of -2 is approximately 0.0228. Multiplying this by the total number of athletes (256), we get 0.0228 * 256 ≈ 5.8368. Since we can't have a fraction of an athlete, the number of junior athletes with heights less than 151 cm is approximately 6.

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• 18.

### The heights of a group of 256 junior athletes is approximately normally distributed with a mean of 157 cm and a standard deviation of 3 cm.   The percentage of the junior athletes with heights between 148 and 166 cm is:

• A.

0.03%

• B.

50%

• C.

68%

• D.

95%

• E.

99.7%

• F.

I don't know

E. 99.7%
Explanation
The answer is 99.7% because in a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, approximately 95% falls within two standard deviations, and approximately 99.7% falls within three standard deviations. Since the range given in the question is within three standard deviations of the mean (148 to 166 cm), it can be concluded that approximately 99.7% of the junior athletes have heights within this range.

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• 19.

### In a normal distribution, approximately 32% of values lie:

• A.

Within one Standard Deviation of the mean

• B.

Within two Standard Deviations of the mean

• C.

Within three Standard Deviations of the mean

• D.

More than one Standard Deviation above or below the mean

• E.

More than two Standard Deviations above or below the mean

• F.

I don't know

D. More than one Standard Deviation above or below the mean
Explanation
In a normal distribution, approximately 68% of values lie within one standard deviation of the mean. Therefore, the remaining approximately 32% of values must lie outside of one standard deviation above or below the mean.

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• 20.

### In a normal distribution, approximately 2.5% of values lie:

• A.

Within one Standard Deviation of the mean

• B.

Within two Standard Deviations of the mean

• C.

Within three Standard Deviations of the mean

• D.

More than one Standard Deviation above the mean

• E.

More than two Standard Deviations above the mean

• F.

I don't know

E. More than two Standard Deviations above the mean
Explanation
In a normal distribution, approximately 2.5% of values lie more than two standard deviations above the mean. This is because in a normal distribution, 95% of values fall within two standard deviations of the mean, leaving only 2.5% of values to be distributed beyond two standard deviations above the mean.

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• 21.

### In a normal distribution, approximately 0.3% of values lie:

• A.

Within one Standard Deviation of the mean

• B.

Within two Standard Deviations of the mean

• C.

Within three Standard Deviations of the mean

• D.

More than three Standard Deviations above or below the mean

• E.

More than two Standard Deviations above or below the mean

• F.

I don't know

D. More than three Standard Deviations above or below the mean
Explanation
In a normal distribution, approximately 0.3% of values lie more than three standard deviations above or below the mean. This is because in a normal distribution, the majority of values are concentrated around the mean, with fewer values as you move further away from the mean. The three standard deviations above or below the mean represent the extreme ends of the distribution, where only a very small percentage of values fall. Therefore, it is expected that approximately 0.3% of values lie in this range.

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• 22.

### In a normal distribution, approximately 95% of values lie:

• A.

Within one Standard Deviation of the mean

• B.

Within two Standard Deviations of the mean

• C.

Within three Standard Deviations of the mean

• D.

More than one Standard Deviation above the mean

• E.

More than two Standard Deviations below the mean

• F.

I don't know

B. Within two Standard Deviations of the mean
Explanation
In a normal distribution, approximately 95% of values lie within two standard deviations of the mean. This means that the majority of the data points fall within a range that is two times the standard deviation above and below the mean. This is a commonly used rule in statistics known as the 95% confidence interval. It indicates that the data is relatively spread out but still follows a bell-shaped curve around the mean.

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• 23.

### A student’s standardised score on a test is –0.5. The mean mark for their class is 68 and the standard deviation is 4.  Their test score is:

• A.

60

• B.

64

• C.

66

• D.

67.5

• E.

70

• F.

I don't know

C. 66
Explanation
The student's standardized score of -0.5 means that their test score is 0.5 standard deviations below the mean mark of 68. Since the standard deviation is 4, 0.5 standard deviations below the mean would be 2 points below the mean. Therefore, the student's test score would be 66.

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• 24.

### A student’s mark on a test is 75. The mean mark for their class is 68 and the standard deviation is 4.  Their standardised score is:

• A.

–2.5

• B.

–1.75

• C.

0

• D.

1.75

• E.

2.5

• F.

I don't know

D. 1.75
Explanation
The student's standardized score is 1.75. Standardized scores, also known as z-scores, measure how many standard deviations a data point is from the mean. In this case, the student's mark of 75 is 1.75 standard deviations above the mean mark of 68. This indicates that the student performed better than the average student in the class.

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• 25.

### It is reasonable to use the mean measure of the centre of a distribution:

• A.

When the distribution is negatively skewed

• B.

When the distribution is positively skewed

• C.

When the distribution is symmetric

• D.

When the distribution is symmetric with outliers

• E.

Always

• F.

I don't know

C. When the distribution is symmetric
Explanation
When the distribution is symmetric, it means that the data is evenly distributed around the mean. In this case, the mean is a good measure of the center because it represents the average value of the data. The mean takes into account all the values in the distribution and provides a balanced representation of the data. However, it may not be the best measure of center when the distribution is skewed or has outliers, as these can heavily influence the mean and make it less representative of the overall data.

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• 26.

### It would not be appropriate to determine the mean and standard deviation of a group of people’s:

• A.

Salary

• B.

Thigh length

• C.

Years of schooling

• D.

School type

• E.

Number of hours worked each week

• F.

I don't know

D. School type
Explanation
Determining the mean and standard deviation of a categorical variable like school type does not provide meaningful information since it is not a numerical variable. Mean and standard deviation are statistical measures used to analyze numerical data, such as salary, thigh length, years of schooling, and number of hours worked each week. Categorical variables can be analyzed using other statistical measures like mode or frequency distribution.

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• 27.

### The following is a set of measurements: 11.0, 11.4, 12.3, 10.5, 11.6, 11.2, 11.8, 11.1, 11.2, 11.3, 11.5   Correct to two decimal places, the actual value of the standard deviation is:

• A.

0.42

• B.

0.44

• C.

0.46

• D.

0.48

• E.

0.50

• F.

I don't know

C. 0.46
Explanation
The correct answer is 0.46 because the standard deviation is a measure of the amount of variation or dispersion in a set of values. To calculate the standard deviation, we need to find the mean (average) of the set of measurements, which is 11.27. Then, for each measurement, we subtract the mean and square the result. We sum up all the squared differences, divide by the number of measurements (n), and take the square root of the result. In this case, the standard deviation is approximately 0.46, which represents the average amount by which each measurement differs from the mean.

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• 28.

### The following is a set of measurements: 11.0, 11.4, 12.3, 10.5, 11.6, 11.2, 11.8, 11.1, 11.2, 11.3, 11.5   An estimate of the standard deviation (based on the range) is:

• A.

0.10

• B.

0.25

• C.

0.45

• D.

0.60

• E.

0.90

• F.

I don't know

C. 0.45
Explanation
The estimate of the standard deviation (based on the range) is 0.45. The range is calculated by subtracting the smallest value from the largest value in the set of measurements, which in this case is 12.3 - 10.5 = 1.8. The estimate of the standard deviation is then obtained by dividing the range by 4, which gives 1.8/4 = 0.45. This estimate provides a rough measure of the spread or variability of the data set.

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• 29.

### The mean of a data distribution is best described as:

• A.

The average

• B.

The middle value

• C.

The central value

• D.

The most common value

• E.

The middle 50% of values

• F.

I don't know

A. The average
Explanation
The mean of a data distribution is best described as "the average". The mean is calculated by summing up all the values in the data set and dividing it by the total number of values. It represents the central tendency of the data and provides a measure of the typical value.

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• 30.

### At Cassie’s school the students are given both a mark and a standardised score for each test they sit. In some recent revision tests, Cassie got the results shown. Assuming the marks are approximately normally distributed, use the standardised scores to decide which of the following statements is not true.

• A.

Cassie’s score in Further Maths is close to the average score for that subject.

• B.

Cassie’s score in English placed her in the top 2.5% of students sitting the test.

• C.

Cassie’s score in Psychology placed her in the bottom 16% of students sitting the test.

• D.

Cassie scored above average in all subjects.

• E.

More than half the class had a higher mark in Biology than Cassie.

• F.

I don't know

D. Cassie scored above average in all subjects.
Explanation
The statement "Cassie scored above average in all subjects" is not true because her score in Psychology placed her in the bottom 16% of students sitting the test, indicating that her score was below average in that subject.

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