Question

Suppose that the average number of Facebook friends users have is normally distributed with a mean of 131 and a standard deviation of about 49. Assume seven 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 7, find the probability that the average number of friends is less than 150.
3. Find the first 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 = = 131

standard deviation = = 49

n = 7

= 131

= / n = 49 / 7  = 18.5203

The distribution of ¯xx¯? ¯xx¯ ~ N( 131 , 18.5203 )

b..

P( < 150 )

= P(( - ) / < ( 150 - 131) / 18.5203 )

= P(z < 1.03 )

Using z table

= 0.8485

Probability = 0.8485

c.

he z dist'n First quartile is,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.6745 ) = 0.25

z = -0.6745

Using z-score formula,

= z * +

= -0.6745 * 18.5203 + 131

= 118.51

The First quartile =Q1 = 118.51

d. Answer = Yes