A survey reports that 53% of college students use Google as their search engine. If 100 students of a college were randomly selected and were asked whether he or she uses Google as their search engine. Using normal approximation to probability, what is the probability (with correction for continuity) that at least 66 will say yes?
We have given,
n=100
p=0.53
np=53
nq=47
Since,
therefore, we need to use here,Normal Approximation for the
Binomial Distribution
=P[X>65.5]
=P[Z>2.5045]
=1-0.9939..................by using Z table.
=0.0061
Therefore, the probability (with correction for continuity) that at least 66 will say yes = 0.0061 |
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