In a survey at a local college, 18% of students reported that they do not own a

ID: 3216584 • Letter: I

Question

In a survey at a local college, 18% of students reported that they do not own a mobile device with internet capabilities. If 200 students are selected at random, find the probability that exactly 50 students don't own a mobile device with internet capabilities.

A magazine reported that 8% of American drivers read the newspaper while driving. If 500 drivers are selected at random, find the probability that more than 38 will admit to reading the newspaper while driving.

Can I please get hope with this 2 questions step by step, thank you

Explanation / Answer

1 ans ) this is a Binomial Distribution, ie a series of n=200 trials with a yes/no outcome for which the probability of a yes is p=0.18

In n trials with a success chance of p, the mean is np and the variance is np(1-p).

n= 200 p= 0.18 q = 0.82

The mean is np=,200*0.18= 36 nq= 200 *0.82 = 164 and

the standard deviation is sqrt(200*0.18*0.82)=5.433

An outcome of exactly 50 yeses corresponds to a range of 49.5 to 50.5,

for which the normalized deviations are
z1=(49.5 - 36)/5.43=2.486
z2=(50.5 - 36)/5.43=2.670

the area between two z values is 0.9962 -0.9934=0.028

so answer is 2.8 %

2ans ) n=500, p=0.08 q = 0.92
mean= 500*0.08= 40 nq = 500*0.92 =460 standard deviation =sqrt(500*0.92*0.08)=6.066

We now set up a Z-score. for 38 or more

Z = (X - mean)/var = (38-40)/6.06 = -0.3303

Looking this up in a standard normal chart we find the probability associated with this Z score is about 0..3703

3.7 %

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In a survey at a local college, 18% of students reported that they do not own a

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