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Lab 03-Lab Ticket Part 2" due Monday, Feb 11 at the beginning of Lab Part 1: Review of HW 3 and Lab 3 (Complete after HW 3 is due: attempt to do a first pass without additional help (only formula card and calculator) Facebook Friends: Based on a study of U.S. Facebook users in 2014, the most connected age group were Facebook users between 18 and 24 years, averaging 649 friends on the social media website. Assume the distribution for the number of Facebook friends for this age group is approximately normal with mean of 649 friends and a standard deviation of 150 friends. Based on the model, determine the value that would correctly complete the sentence below The top 15% of most connected Facebook users between 18 to 24 years will have at least a. riends Final Answer: at least friends Based on the model, what is the probability that a randomly selected Facebook user between 18 to 24 years has at least 844 friends? b. Final Answer Consider a random sample of 8 Facebook users between 18 to 24 years. What is the probability that exactly 3 of the 8 Facebook users will have at least 844 friends? c. Final Answer:

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Sean

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Step 1/4

The provided information is as follows: The distribution of the number of Facebook friends on the social media website is approximately normal. The mean is `mu=649` . The standard deviation is `sigma=150` .

Step 2/4

The number of Facebook friends for the age group 18 and 24 years is `X` . Here, `X"~Normal(649, 150)"` . a) The least number of friends whose top 15% of most connected Facebook users between 18 to 24 years is calculated as follows:

`P(Xgex) = 0.15`

`1-P(X<x) = 0.15`

`P(X<x) = 1-0.15`

`P(X<x) = 0.85`

`P((X-mu)/sigma<(x-649)/150) = 0.85`

`P(Z<(x-649)/150) = 0.85`

`(x-649)/150 = phi^(-1)(0.85)`

`x = 150*phi^(-1)(0.85)+649`

` = 150*1.0364+649`

` \cong 804.46`

` \cong 805`

The least number of friends whose top 15% of most connected Facebook users between 18 to 24 years is 805.

Explanation:

The formula for the Z-score is `Z=(X-mu)/sigma` .

Explanation:

The required value is calculated by using the Excel formula `phi^(-1)(0.85)=NORM.S.INV(0.85)` .

Step 3/4

b) The probability that a randomly selected Facebook user between 18 and 24 years has at least 844 friends is calculated as follows:

`P(Xge844) = 1-P(X<844)`

` = 1-P((X-mu)/sigma<(844-649)/150)`

` = 1-P(Z<1.3)`

` \cong 1-0.9032`

` = 0.0968`

The probability that a randomly selected Facebook user between 18 and 24 years has at least 844 friends is 0.0968 .

Explanation:

The required probability is calculated using the Excel formula `P(Z<z)=NORM.S.DIST(z,TRUE)` which is `P(Z<1.3)=NORM.S.DIST(1.3,TRUE)`

Step 4/4

c) Here, the distribution is binomial with `(8,0.0968)` .

Explanation:

Here, we are interested in the probability that exactly 3 out of 8 users have at least 844 friends, and the binomial distribution is a suitable model for this scenario as the total number of trials which is the total number of Facebook users are fixed, each trial has two possible outcomes either number of users having at least 844 friends or not and the probability of success remains constant and each trial, therefore distribution is binomial.

The probability that exactly 3 of the 8 users will have at least 844 friends is calculated as follows

`P(X=3) = ^8C_3*0.0968^3*(1-0.0968)^(8-3)`

` = (8!)/(3!(8-3)!)*0.0005451872824`

` = (8*7*6*5*4*3*2*1)/(3*2*1*5*4*3*2*1)*0.0005451872824`

` \cong 0.0305`

The probability that exactly 3 of the 8 users will have at least 844 friends is 0.0305 .

Explanation:

The probability formula for the binomial distribution is `P(X=x)=^nC_x*p^x*(1-p)^(n-x)` where `^nC_X=(n!)/(x!(n-x)!` and `n! =n*(n-1)*.....*1`

Final Answer

a) The least number of friends whose top 15% of most connected Facebook users between 18 to 24 years is 805. b) The probability that a randomly selected Facebook user between 18 and 24 years has at least 844 friends is 0.0968 . c) The probability that exactly 3 of the 8 users will have at least 844 friends is 0.0305 .

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