Question

Suppose 5% of college students don’t use social media. We randomly sampled 1000 college students.

a. Let X denote the number of students who don’t use social media in this sample. Calculate the expected value and the standard deviation for X.

b. Find the probability of observing 50 or less students who don’t use social media in this sample. If you would like to use normal approximation for binomial distribution, make sure to check the conditions and you don’t need to do the continuity correction.

Answer #1

a) we are given

n = sample size = 1000

p = proportion = 0.05

Let X : number of students who don’t use social media follow the binomial distribution

So expected value E( X ) = n * p = 1000*0.05

= E( X ) = 50

and standard deviation = sqrt [ n * p* q] = sqrt [ 1000*0.05*0.85]

Standard deviation = 6.89

b) Normal approximation to binomial distribution

n * p = 50 > 10 and n *( 1 -p) = 950 > 10

Both conditions are satisfied

P( X 50 ) = P ( X - np/sqrt(n*p*q) < 50 - np/sqrt (n*p*q)]

P( X 50 ) = P( Z 0)

P( X 50) = 0.5000

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