Question

A survey was conducted to determine the number of cats living in individual households. The results are summarized in the accompanying table. Complete parts a and b.

Number of Cats	Probability
0	0.27
1	0.26
2	0.28
3	0.15
4	0.04

Determine the mean and standard deviation. (Do not round)

Answer 1

Solution :

x	P(x)	x * P(x)	x2 * P(x)
0	0.27	0	0
1	0.26	0.26	0.26
2	0.28	0.56	1.12
3	0.15	0.45	1.35
4	0.04	0.16	0.64
Sum	1	1.43	3.37

Mean = = X * P(X) = 1.43

Standard deviation =

=X ² * P(X) - ²

=  3.37 - 1.43²

^{= 1.15113}