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.
to illustrate the importance of outliers, suppose the 65-year old walks away from the party. remaining at the party are the five people who are 30 years old, three people that are 32, and one person that is 31.
determine the mean age and standard deviation, to the nearest tenth, of the people who remain at the party.
a. the mean age of the remaining people at the party is a0
b. the standard deviation is:

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

The first answer is correct