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• RD Sharma Class 10 Solutions

# Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.3 | Set 1

### Find the average expenditure (in rupees) per household.

Solution:

Let the assumed mean (A) = 275

It’s seen that A = 275 and h = 50

So,

Mean = A + h x (Σfi ui/N)

= 275 + 50 (-35/200)

= 275 – 8.75

= 266.25

### Which method did you use for finding the mean, and why?

Solution:

From the given data,

To find the class interval we know that,

Class marks (xi) = (upper class limit + lower class limit)/2

Now, let’s compute xi and fixi by the following

Here,

Mean = Σ fiui/N

= 162/ 20

= 8.1

Thus, the mean number of plants in a house is 8.1

We have used the direct method as the values of class mark xi and fi is very small.

### Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution:

Let us assume mean (A) = 150

It’s seen that,

A = 150 and h = 20

So,

Mean = A + h x (Σfi ui/N)

= 150 + 20 x (-12/50)

= 150 – 24/5

= 150 = 4.8

= 145.20

### Question 4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Solution:

Using the relation (xi) = (upper class limit + lower class limit)/ 2

And, class size of this data = 3

Let the assumed mean (A) = 75.5

So, let’s calculate di, ui, fiui as following:

From table, it’s seen that

N = 30 and h = 3

So, the mean = A + h x (Σfi ui/N)

= 75.5 + 3 x (4/30

= 75.5 + 2/5

= 75.9

Therefore, the mean heart beats per minute for those women are 75.9 beats per minute.

### Question 5. Find the mean of each of the following frequency distributions:

Solution:

Let’s consider the assumed mean (A) = 15

From the table it’s seen that,

A = 15 and h = 6

Mean = A + h x (Σfi ui/N)

= 15 + 6 x (3/40)

= 15 + 0.45

= 15.45

### Question 6. Find the mean of the following frequency distribution:

Solution:

Let’s consider the assumed mean (A) = 100

From the table it’s seen that,

A = 100 and h = 20

Mean = A + h x (Σfi ui/N)

= 100 + 20 x (61/100)

= 100 + 12.2

= 112.2

### Question 7. Find the mean of the following frequency distribution:

Solution:

From the table it’s seen that,

A = 20 and h = 8

Mean = A + h x (Σfi ui/N)

= 20 + 8 x (7/40)

= 20 + 1.4

= 21.4

### Question 8. Find the mean of the following frequency distribution:

Solution:

Let’s consider the assumed mean (A) = 15

From the table it’s seen that,

A = 15 and h = 6

Mean = A + h x (Σfi ui/N)

= 15 + 6 x (5/40)

= 15 + 0.75

= 15.75

### Question 9. Find the mean of the following frequency distribution:

Solution:

Let’s consider the assumed mean (A) = 25

From the table it’s seen that,

A = 25 and h = 10

Mean = A + h x (Σfi ui/N)

= 25 + 10 x (8/60)

= 25 + 4/3

= 79/3 = 26.333

### Question 10. Find the mean of the following frequency distribution:

Solution:

Let’s consider the assumed mean (A) = 20

From the table it’s seen that,

A = 20 and h = 8

Mean = A + h x (Σfi ui/N)

= 20 + 8 x (5/40)

= 20 + 1

= 21

### Question 11. Find the mean of the following frequency distribution:

Solution:

Let’s consider the assumed mean (A) = 20

From the table it’s seen that,

A = 20 and h = 8

Mean = A + h x (Σfi ui/N)

= 20 + 6 x (-9/20)

= 20 – 72/20

= 20 – 3.6

= 16.4

### Question 12. Find the mean of the following frequency distribution:

Solution:

Let’s consider the assumed mean (A) = 60

From the table it’s seen that,

A = 60 and h = 20

Mean = A + h x (Σfi ui/N)

= 60 + 20 x (14/50)

= 60 + 28/5

= 60 + 5.6

= 65.6

### Question 13. Find the mean of the following frequency distribution:

Solution:

Let’s consider the assumed mean (A) = 50

From the table it’s seen that,

A = 50 and h = 10

Mean = A + h x (Σfi ui/N)

= 50 + 10 x (-2/40)

= 50 – 0.5

= 49.5

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