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 = 30 years old

3 people = 32 years old

1 person = 31 years old

1 person = 65 years old

We need to arrange the ages in order from least to greatest.

30, 30, 30, 30, 30, 31, 32, 32, 32, 65

65 is the outlier.

A.) mean age without the outlier. count of the numbers : 9

5 x 30 = 150

1 x 31 = 31

3 x 32 = 96

total age 277

: # of pax 9

mean 30.77 round to the nearest tenth 30.80

b.) Standard deviation:

We need to calculate variance first:

5 pax : 30 - 30.80 = - 0.80 ⇒ - 0.80² = 0.64 * 5pax = 3.20

1 pax : 31 - 30.80 = 0.20 ⇒ 0.20² = 0.04 * 1pax = 0.04

3 pax : 32 - 30.80 = 1.20 ⇒ 1.20² = 1.44 * 3 pax = 4.32

Variance : (3.20 + 0.04 + 4.32) / 9 pax = 7.56 / 9 = 0.84

Standard deviation is simply the square root of the variance.

Standard deviation = √0.84 = 0.916

0.916 round to nearest hundredths is 0.92

0.916 round to nearest tenths is 0.90