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# Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.2

• Difficulty Level : Easy
• Last Updated : 21 Dec, 2020

### Compute the mean number of calls per interval.

Solution:

Let the assumed mean(A) be =3 (Generally we choose the middle element to be the assumed mean, but it’s not mandatory),

hence, the table is,

hence, mean of the calls = = = Therefore, mean number of calls per interval is 3.54

### Question 2: Five coins were simultaneously tossed 1000 times, and at each toss the number of heads was observed. The number of tosses during which 0, 1, 2, 3, 4, and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.

Solution:

Let the assumed mean (A) be = 2

hence, the table is,

Mean number of head per toss = = = 2.47

Therefore, mean number of head per toss is 2.47

### Calculate the average number of branches per plant.

Solution:

Let the assumed mean(A) be = 4

hence, the table is,

Average Number of branches per plant = = = 3.615

Therefore, mean number of branches per plant is 3.615

### Find the average number of children per family.

Solution:

Let the assumed mean(A) be = 2

Hence, the table is,

Average number of children per family = = = 2.35 (approximately)

Therefore, the average number of children per family is 2.35 (approximately)

### Find the average number of marks.

Solution:

Let the assume mean (A) be = 25

hence, the table is,

Average number of marks = = = 26.08 (approximately)

Therefore, average number of marks is 26.08 (approximately)

### Find the mean number of students absent per day.

Solution:

Let mean assumed mean (A) be = 3

Mean number of students absent per day = = = 3.525

Therefore, the mean number of students absent per day is 3.525

### Find the average number of misprints per page.

Solution:

Let the assumed mean (A) be = 2

Average number of misprints per day = = = 0.73

Therefore, the average number of misprints per day is 0.73

### Find the average number of misprints per page.

Solution:

Let the assumed mean (A) = 2

Average no of accidents per day workers = = = 0.83

Therefore, average no of accidents per day workers 0.83

### Question 9: Find the mean from the following frequency distribution of marks at a test in statistics:

Solution:

Let the assumed mean (A) be = 25

Mean = = = 22.075

Therefore, the mean is 22.075

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