Question

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

p=5% =0.05

q=1-p = 1-0.05 =0.95

n=1000

Since,n is large,so we can use normal approximation for binomial distribution

Expected value = μ=n*p = 1000*0.05 =**50**

standard deviation = σ=√(npq) = √(1000*0.05*0.95)

=**6.892**

x=50

z=(x-μ)/σ

=(50-50)/6.892

=0

P(x<50) = P(z<0)

=
**0.5** {using normal distribution table}

OR

Using Excel formula of Binomial distribution

we do =BINOM.DIST(50,1000,0.05,TRUE)

Thus,we get **0.537529041**

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