Question

Suppose that the average number of Facebook friends users have is normally distributed with a mean of 119 and a standard deviation of about 41. Assume forty-five individuals are randomly chosen. Answer the following questions. Round all answers to 4 decimal places where possible.

1. What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)
2. For the group of 45, find the probability that the average number of friends is more than 114.
3. Find the third quartile for the average number of Facebook friends.
4. For part b), is the assumption that the distribution is normal necessary? NoYes

Answer 1

Solution :

Given that ,

mean = = 119

standard deviation = = 41

n = 45

a) =   = 119

= / n = 41 / 45 = 6.11

The sampling distributiion of is approximately normal   N ( 119, 6.11)

b) P( > 114) = 1 - P( < 114)

= 1 - P[( - ) / < (114 - 119) / ]

= 1 - P(z < -0.82)   

= 1 - 0.2061

= 0.7939

c) The z dist'n Third quartile is,

= P(Z < 0.674 ) = 0.75

z = 0.674

Using z-score formula,

= z * +

= 0.674 * 6.11 + 119

= 123.12

Third quartile =Q3 = 123.12

d) yes