The mean age is 34.2. I'm working on the standard deviation right now, I'll comment on this post when I figure it out.
Answer:
A. The mean age of the ten people at a party is: 34.2
B. The standard deviation is: 10.3066
Step-by-step explanation:
We are given a information about the age of ten people as:
Age(in years) Number of people
30 5
32 3
31 1
65 1
Hence, the mean of the age is calculated as:
[tex]Mean(x')=\dfrac{30\times 5+32\times 2+31+65}{10}\\\\\\Mean(x')=\dfrac{342}{10}\\\\\\Mean(x')=34.2[/tex]
Now, we calculate the standard deviation in the following steps:
x x-x' (x-x')²
30 -4.2 17.64
30 -4.2 17.64
30 -4.2 17.64
30 -4.2 17.64
30 -4.2 17.64
32 -2.2 5.06
32 -2.2 5.06
32 -2.2 5.06
31 -3.2 10.24
65 30.8 948.64
∑ (x-x')²= 1062.26
Variance=∑ (x-x')²/10=106.226
Hence, standard deviation is the square root of variance.
standard deviation is:
[tex]=\sqrt{106.226}=10.3066[/tex]
