CBSE Class 9 Maths Chapter 14 Statistics and Probability NCERT Solutions

To find the class mark of a given class interval, we need to calculate the average of its lower and upper limits. The formula for the class mark is:

$$Class\; mark\; =\; \backslash frac\{Upper\backslash \; limit\; +\; Lower\backslash \; limit\}\{2\}$$

In this case, the lower limit is 90 and the upper limit is 120.

Substituting these values into the formula:

$$Class\; mark\; =\; \backslash frac\{120\; +\; 90\}\{2\}$$

$$Class\; mark\; =\; \backslash frac\{210\}\{2\}$$

$$Class\; mark\; =\; 105$$

Therefore, the class mark of the class 90-120 is 105. The correct option is (B).

The range of a data set is defined as the difference between the maximum value and the minimum value present in the data set. First, we need to identify the highest and lowest values in the given data.

The given data set is: 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20.

The minimum value in this data set is 6.

The maximum value in this data set is 32.

Now, we calculate the range using the formula:

$$Range\; =\; Maximum\; value\; -\; Minimum\; value$$

Substituting the identified values:

$$Range\; =\; 32\; -\; 6$$

$$Range\; =\; 26$$

Thus, the range of the given data is 26. The correct option is (D).

We are given the mid-value of a class and the width of the class. Let the upper class limit be denoted by $U$ and the lower class limit be denoted by $L$.

We know the following formulas:

1. Mid-value = $$\backslash frac\{U\; +\; L\}\{2\}$$
2. Class Width = $U - L$

Given:

• Mid-value = 10
• Class Width = 6

Using the mid-value formula:

$$10\; =\; \backslash frac\{U\; +\; L\}\{2\}$$

Multiplying both sides by 2, we get:

$$20\; =\; U\; +\; L$$ ... (Equation 1)

Using the class width formula:

$$6\; =\; U\; -\; L$$ ... (Equation 2)

Now, we can solve these two linear equations for $U$ and $L$. To find $L$, we can subtract Equation 2 from Equation 1:

(Equation 1) - (Equation 2):

$$(U\; +\; L)\; -\; (U\; -\; L)\; =\; 20\; -\; 6$$

$$U\; +\; L\; -\; U\; +\; L\; =\; 14$$

$$2L\; =\; 14$$

Dividing by 2:

$$L\; =\; \backslash frac\{14\}\{2\}$$

$$L\; =\; 7$$

So, the lower limit of the class is 7. The correct option is (B).

(We can also find the upper limit by adding Equation 1 and Equation 2: $$2U\; =\; 26\; \backslash implies\; U\; =\; 13$$. The class interval is 7-13, with a mid-value of (7+13)/2 = 10 and a width of 13-7 = 6.)

We are given that there are five continuous classes, each with a width of 5. The lower class-limit of the very first (lowest) class is 10.

Let's list the class intervals:

• Class 1: The lower limit is 10. Since the width is 5, the upper limit is $10 + 5 = 15$. So, the first class is 10-15.
• Class 2: The lower limit is the same as the upper limit of the previous class, which is 15. The width is 5, so the upper limit is $15 + 5 = 20$. The second class is 15-20.
• Class 3: The lower limit is 20. The upper limit is $20 + 5 = 25$. The third class is 20-25.
• Class 4: The lower limit is 25. The upper limit is $25 + 5 = 30$. The fourth class is 25-30.
• Class 5: The lower limit is 30. The upper limit is $30 + 5 = 35$. The fifth class is 30-35.

We have listed all five continuous classes. The highest class is the fifth class, which is 30-35.

The upper class-limit of the highest class (30-35) is 35.

Alternatively, we can calculate this directly. The total width covered by 5 classes is $5 \times 5 = 25$. The upper limit of the highest class will be the lower limit of the lowest class plus the total width. Upper limit of highest class = Lower limit of lowest class + (Number of classes $\times$ Class width) = $10 + (5 \times 5) = 10 + 25 = 35$.

Therefore, the upper class-limit of the highest class is 35. The correct option is (C).