Consider the probability that greater than 90 out of 149 software users will not call technical support. Assume the probability
1 answer:
Answer:
0.1220
Step-by-step explanation:
You want the probability that greater than 90 out of 149 software users will not call tech support, given that the probability a given user will not call is 56%. You want the value using the normal approximation to the binomial distribution.
<h3>Normal approximation</h3>For a sample size of n, and probability p of "success", the normal approximation to the binomial distribution can be used if ...
min(pn, (1-p)n) > 5
Here, that is ...
min(0.56·149, 0.44·149) > 5
65.6 > 5 . . . . . . . . true, the normal approximation is appropriate
Then the parameters of the normal distribution are ...
µ = pn = 0.56·149 = 83.44
σ = √((1-p)µ) = √(0.44·83.44) ≈ 6.0592
<h3>Continuity correction factor</h3>The probability that X = N will occur can be considered to be ...
P(N -0.5 < X < N +0.5)
Then the probability that X > N will be modeled using the normal distribution as ...
P(X > N +0.5)
For this problem, we want ...
P(X > 90.5) . . . . using the normal distribution with µ = 83.44, σ = 6.0592
The attached calculator shows that value to be ...
P(>90) ≈ 0.1220 . . . . . . rounded to 4 decimal places
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<em>Additional comment</em>
A probability calculator says the actual binomial distribution result for this problem is 0.1217.
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