Question

Ten people at a party decide to compare ages. Five people are 30 years old, three are 32, one is 31, and one is 65.

Answer 1

Given:
Ten people:
5 people are 30 years old
3 people are 32 years old
1 person is 31 years old
1 person is 65 years old ⇒Outlier. No included in the next computation.

A.) Mean Age of the remaining 9 people.

Mean = (5(30) + 3(32) + 31) / 9
          = (150 + 96 + 31)/9
          = 277/9
          = 30.77 or 30.80 is the mean age to the nearest tenth.

We need to get the variance before we can solve the standard deviation.

Variance: 
1 ⇒ 30 - 30.80 = -0.80 ⇒ -0.80² = 0.64
2 ⇒ 30 - 30.80 = -0.80 ⇒ -0.80² = 0.64
3 ⇒ 30 - 30.80 = -0.80 ⇒ -0.80² = 0.64 
4 ⇒ 30 - 30.80 = -0.80 ⇒ -0.80² = 0.64
5 ⇒ 30 - 30.80 = -0.80 ⇒ -0.80² = 0.64
6 ⇒ 31 - 30.80 =  0.20 ⇒  0.20² = 0.04
7 ⇒ 32 - 30.80 =  1.20 ⇒  1.20² = 1.44
8 ⇒ 32 - 30.80 =  1.20 ⇒  1.20² = 1.44
9 ⇒ 32 - 30.80 =  1.20 ⇒  1.20² = 1.44
                                   sum:           7.56
                           divide by:                9
                            quotient:           0.84 Variance

Standard Deviation is the square root of the Variance.

Standard Deviation = √0.84  = 0.92