Question

The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then which one of the following gives possible values of a and b?

• A. a = 0, b = 7
• B. a = 5, b = 2
• C. a = 3, b = 4
• D. a = 2, b = 4

Answer 1

The Correct Option is C

To find the possible values of a and b that satisfy the conditions given, we must use the formulas for mean and variance and the provided data.
First, calculate the mean of the numbers: The mean (\(\mu\)) is given as 6. The formula is:

\[\mu = \frac{a+b+8+5+10}{5}\]
Simplifying, we have:
\[ \frac{a+b+23}{5} = 6\]
Multiplying both sides by 5 gives:
\[ a+b+23 = 30\]
Subtract 23 from both sides:
\[ a+b = 7\]
Next, the variance (\(\sigma^2\)) is given as 6.80. The formula for variance is:

\[\sigma^2 = \frac{(a-\mu)^2 + (b-\mu)^2 + (8-\mu)^2 + (5-\mu)^2 + (10-\mu)^2}{5}\]
Substitute \(\mu = 6\):

\[6.80 = \frac{(a-6)^2 + (b-6)^2 + (8-6)^2 + (5-6)^2 + (10-6)^2}{5}\]
Calculate the squared differences for known numbers:

• (8-6)^2 = 4
• (5-6)^2 = 1
• (10-6)^2 = 16

Substitute these values:
\[6.80 = \frac{(a-6)^2 + (b-6)^2 + 4 + 1 + 16}{5}\]
\[6.80 = \frac{(a-6)^2 + (b-6)^2 + 21}{5}\]
Multiply through by 5 and simplify:
\[34 = (a-6)^2 + (b-6)^2 + 21\]
Subtract 21 from both sides:
\[13 = (a-6)^2 + (b-6)^2\]
We know \(a+b=7\). Consider the options:

a=0, b=7	\((0-6)^2 + (7-6)^2 = 36 + 1 = 37\)
a=5, b=2	\((5-6)^2 + (2-6)^2 = 1 + 16 = 17\)
a=3, b=4	\((3-6)^2 + (4-6)^2 = 9 + 4 = 13\)
a=2, b=4	\((2-6)^2 + (4-6)^2 = 16 + 4 = 20\)

The correct pair is \(a=3\) and \(b=4\) because it satisfies both the mean and variance conditions.