The mode score on the 6th grade math test was 94. which of the following interpretation is correct

Mean, Median and Mode are three measures of central tendency, and these are the important topics in statistics. In this article, you will get mean, median, and mode questions and answers, along with some practice questions. These questions and answers will help you to practise for the board exams. Also, these questions cover various types of problems asked on mean, median and mode for Classes 9 and 10.

What is the mean formula?

As we know, the mean is the average of the given data set. This can be calculated using the formula given below.

Mean for ungrouped data = Sum of observations/Number of observations

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What is the formula for the median?

The median is the middlemost data value of an ordered data set. This can be estimated using the formulas given below.

When n is odd:

Median = (n + 1)/2th observation

When n is even:

Median = [(n/2)th observation + {(n/2)+1}th observation]/2

Learn more about median here.

What is the mode?

Mode is the most frequently occurring value in the data set.

Get more information about mode here.

Mean Median Mode Questions and Answers

1. Find the mean of the first 10 odd integers.

Solution:

First 10 odd integers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Mean = Sum of the first 10 odd integers/Number of such integers

= (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)/10

= 100/10

= 10

Therefore, the mean of the first 10 odd integers is 10.

2. What is the median of the following data set?

32, 6, 21, 10, 8, 11, 12, 36, 17, 16, 15, 18, 40, 24, 21, 23, 24, 24, 29, 16, 32, 31, 10, 30, 35, 32, 18, 39, 12, 20

Solution:

The ascending order of the given data set is:

6, 8, 10, 10, 11, 12, 12, 15, 16, 16, 17, 18, 18, 20, 21, 21, 23, 24, 24, 24, 29, 30, 31, 32, 32, 32, 35, 36, 39, 40

Number of values in the data set = n = 30

n/2 = 30/2 = 15

15th data value = 21

(n/2) +1 = 16

16th data value = 21

Median = [(n/2)th observation + {(n/2)+1}th observation]/2

= (15th data value + 16th data value)/2

= (21 + 21)/2

= 21

3. Identify the mode for the following data set:

21, 19, 62, 21, 66, 28, 66, 48, 79, 59, 28, 62, 63, 63, 48, 66, 59, 66, 94, 79, 19 94

Solution:

Let us write the given data set in ascending order as follows:

19, 19, 21, 21, 28, 28, 48, 48, 59, 59, 62, 62, 63, 63, 66, 66, 66, 66, 79, 79, 94, 94

Here, we can observe that the number 66 occurred the maximum number of times.

Thus, the mode of the given data set is 66.

4. Consider the following frequency distribution. Calculate the mean weight of students.

Solution:

The given distribution has discontinuous class intervals, so we need to make them continuous.

Here, ∑fi = 40 and ∑fidi = 35

By Assumed mean method,

Mean = a + (∑fidi/∑fi)

= 43 + (35/40)

= 43 + 0.875

= 43.875

Therefore, the mean weight of the students is 43.875 kg.

5. Find the mean for the following distribution.

Solution:

Mean = ∑fixi/∑fi

= 802/29

= 27.655

6. Calculate the median marks of students from the following distribution.

Solution:

N/2 = 90/2 = 45

Cumulative frequency greater and nearer to 45 is 47, which lies in the interval 40 – 50

Median class is 40 – 50.

Lower limit of the median class = l = 40

Class size = h = 10

Frequency of the median class = f = 20

Cumulative frequency of the class preceding the median class = cf = 27

As we know,

\(\begin{array}{l}Median = l + \left ( \frac{\frac{N}{2}-cf}{f} \right )\times h\end{array} \)

Median = 40 + [(45 – 27)/20] × 10

= 40 + (18/2)

= 40 + 9

= 49

Hence, the median marks of the students = 49.

7. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the above distribution.

Solution:

From the given,

Modal class = 4000 – 5000

Lower limit of the modal class = l = 4000

Class width (h) = 1000

Frequency of the modal class = f1 = 18,

Frequency of the class preceding the modal class = f0 = 4

Frequency of the class succeeding the modal class = f2 = 9

Mode = l + [(f1 – f0)/ (2f1 – f0 – f2)] × h

Mode = 4000 + [(18 – 4)/(36 – 4 – 9)] × 1000

= 4000 + (14000/23)

= 4000 + 608.695

= 4608.695

= 4608.7 (approximately)

Thus, the mode of the given data is 4608.7 runs

8. If the median of a distribution given below is 28.5, then find the value of x and y.

Solution:

From the given,

N/2 = 60/2 = 30

Median of the given data = 28.5

Median class is 20 – 30 with a cumulative frequency = 25 + x.

Lower limit of median class = l = 20

Frequency of the median class = f = 20

Cumulative frequency of the class preceding the median class = cf = 5 + x

Class size = h = 10

\(\begin{array}{l}Median = l + \left ( \frac{\frac{N}{2}-cf}{f} \right )\times h\end{array} \)

28.5 = 20 + [(30 − 5 − x)/ 20] × 10

28.5 – 20 = (25 – x)/2

8.5 × 2 = 25 – x

17 = 25 – x

x = 25 – 17

Therefore, x = 8

Also,

60 = 5 + 20 + 15 + 5 + x + y

60 = 5 + 20 + 15 + 5 + 8 + y

y = 60 – 53

y = 7

Therefore, the value of x = 8 and y = 7.

9. For a moderately skewed distribution, the mean and median are respectively 26.8 and 27.9. What is the mode of the distribution?

Solution:

Given,

Mean = 26.8

Median = 27.9

Using the relationship between mean, median and mode,

Mode = 3 Median – 2 Mean

= 3 × 27.9 – 2 × 26.8

= 83.7 – 53.6

= 30.1

Therefore, the mode of the distribution is 30.1.

10. If the mean of the given frequency distribution is 35, then find the missing frequency y. Also, calculate the median and mode for the distribution.

Solution:

As we know,

Mean = ∑fixi/∑fi

Given that the mean of the distribution is 35.

So, (430 + 45y)/(14 + y) = 35

430 + 45y = (14 + y)35

430 + 45y = 490 + 35y

45y – 35y = 490 – 430

10y = 60

y = 6

Thus, the missing frequency is 6.

Now, we can calculate the mode as follows:

Here,

l = 30

h = 10

Mode = l + [(f1 – f0)/ (2f1 – f0 – f2)] × h

Mode = 30 + [(7 – 4)/(14 – 4 – 6)] × 10

= 30 + (30/4)

= 30 + 7.5

= 37.5

Now, using the formula Mode = 3Median – 2 Mean, we can get the value of median.

37.5 = 3 Median – 2 (35)

3 Median = 37.5 + 70

Median = 107.7/3

= 35.9

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Practice Questions on Mean Median Mode

1. Find the mean, median and mode of the following data:

23, 18, 24, 23, 31, 37, 28, 30, 25, 40, 35, 35, 27, 25

2. The mileage (km per litre) of 50 cars of the same model was tested by a manufacturer, and details are tabulated as given below:

Find the mean mileage.

The manufacturer claimed that the mileage of the model was 16 km/litre. Do you agree with this claim?

3. Find the median of the following frequency distribution.

4. Calculate the mode for the following distribution.

5. Suppose the mean and median of distribution are 10.14 and 8, respectively. Find the mode of the distribution using the empirical relationship between mean, median and mode.

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