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adell [148]
1 year ago
5

Consider the probability that greater than 90 out of 149 software users will not call technical support. Assume the probability

that a given software user will not call technical support is 56%. Approximate the probability using the normal distribution. Round your answer to four decimal places.
Mathematics
1 answer:
Artemon [7]1 year ago
7 0

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

__

<em>Additional comment</em>

A probability calculator says the actual binomial distribution result for this problem is 0.1217.

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