1. Each item produced by a certain manufacturer is, independently, of acceptable quality with probability $0.95$. Approximate the probability (by a normal distribution) that at most $10$ of the next $150$ items produced are unacceptable.

  2. Hits to a high-volume Web site are assumed to follow a Poisson distribution with a mean of $10,000$ per day.

$\,\,\,\,\,\,\,\,\,\,\text{(a)}$ Approximate (by a normal distribution) the expected number of days in a year ($365$ $\,\,\,\,\,\,\,\,\,\,\,$years) that exceed $10,200$ hits.

$\,\,\,\,\,\,\,\,\,\,\text{(b)}$ Approximate (by a normal distribution) the probability that over a year ($365$ days) $\,\,\,\,\,\,\,\,\,\,\,$more than $15$ days each have more than $10,200$ hits.

For question 1, is the answer $0.8695$?

For question 2(a), is that I have to use P(X<= 10199.5) to find the probability and multiply the probability by $365$ ?

For question 2(b), would anyone mind telling me how to solve it?


For Question 1)

Model the number of unacceptable items as normal, p=.05.

$\hat\mu = .05(150) = 7.5\space,\hat\sigma = \sqrt{150(.05)(.95)} = 2.67$

Normal Approximation: F = No. of Unacceptable Items, $F \sim N(7.5,2.67)$

$P(F\leq 10) = \Phi(\frac{10-7.5+.5}{0.84}) = 0.87$ Using the continuity correction.

So, your first answer is correct.

Question 2)

a) Your approach is generally correct. You want to calcluate $P(X\geq 10,201)$ using the normal approximation ($\hat \mu = \hat \sigma^2 = 10,000 $and multiply by 365.

b) HINT: You'll need to first approximate $P(X\geq 10,201)$ as before then use the Binomial distribution to get the probability for the number of days.

  • $\begingroup$ So, for question 2(b), is the answer 0.9906 ? $\endgroup$ – user100523 Nov 27 '13 at 4:42
  • $\begingroup$ you mean I should use the negative binomial distribution? $\endgroup$ – user100523 Nov 27 '13 at 4:49
  • $\begingroup$ Part a is asking for an expected value not a probability. also yours seems very high. for B....No, binomial dist should be used $\endgroup$ – user76844 Nov 27 '13 at 11:30
  • $\begingroup$ Well, I mean for 2(a), the answer is 8.103 days, so is the answer 9 days? $\endgroup$ – user100523 Nov 27 '13 at 12:38
  • $\begingroup$ Also, for 2(b), is the answer 0.01134 ? $\endgroup$ – user100523 Nov 27 '13 at 12:39

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