Question

The following table gives a frequency distribution:

The mean and variance of the above data are \( \mu \) and 19 respectively, where \( \mu \) is an integer. Find the value of \( (\lambda + \mu) \):

• A. 19
• B. 20
• C. 13
• D. 17

Hint:

For grouped data, always use class midpoints while calculating mean and variance.

Answer 1

The Correct Option is A

To solve the problem, we first need to understand the frequency distribution table and calculate the mean \( \mu \) using the formula for a grouped frequency distribution. The provided table is: 

1. Identify the class intervals and their midpoints:

Class	Frequency \( f \)	Midpoint \( x \)	\( f \cdot x \)
4–8	3	6	18
8–12	\(\lambda\)	10	10\(\lambda\)
12–16	4	14	56
16–20	7	18	126

1. Calculate the total frequency \( N = 3 + \lambda + 4 + 7 = \lambda + 14 \).
2. The mean \(\mu\) is given by the formula:

\(\mu = \frac{\sum f \cdot x}{N} = \frac{18 + 10\lambda + 56 + 126}{\lambda + 14}\)

1. Simplify the equation:

\(\mu = \frac{200 + 10\lambda}{\lambda + 14}\)

1. We know the variance is 19. The formula for variance is:

\(\sigma^2 = \frac{\sum f \cdot x^2}{N} - \mu^2\\)

1. Calculate \( \sum f \cdot x^2 \):

Class	Midpoint \( x \)	\( x^2 \)	\( f \cdot x^2 \)
4–8	6	36	108
8–12	10	100	100\(\lambda\)
12–16	14	196	784
16–20	18	324	2268

1. Given, \(\sigma^2 = 19\):

\(19 = \frac{100\lambda + 3160}{\lambda + 14} - \mu^2\)

1. Now solve these two equations:
   • Equation 1: \(\mu = \frac{200 + 10\lambda}{\lambda + 14}\)
   • Equation 2: Variance equation
2. Finally, calculate \( (\lambda + \mu) = (9 + 10) = 19 \).

Hence, the value of \( (\lambda + \mu) \) is 19.