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According to Boeing data, the 757 airliner carries 200 passengers and has doors with a mean height of 72 inches. Assume for a certain population of men we have a mean of 69.0 inches and a standard deviation of 2.8 inches.

  1. What mean doorway height would allow 95% of men to enter the aircraft without bending?
  2. Assume that half of the 200 passengers are men. What mean doorway height satisfies the condition that there is a 0.95 probability that this height is greater than the mean height of 100 men?
  3. For engineers designing the 757, which result is more relevant: the height from part a or part b? Why?
  1. We know that μ x = μ = 69 and we have σ x = 2.8. The height of the doorway is found to be = 73.61
  2. We know that μ x = μ = 69 and we have σ x = 0.28. So we get 69.49
  3. When designing the doorway heights, we need to incorporate as much variability as possible in order to accommodate as many passengers as possible. Therefore, we need to use the result based on part a.

Historical note

: Normal Approximation to the Binomial

Historically, being able to compute binomial probabilities was one of the most important applications of the central limit theorem. Binomial probabilities with a small value for n (say, 20) were displayed in a table in a book. To calculate the probabilities with large values of n , you had to use the binomial formula, which could be very complicated. Using the normal approximation to the binomial distribution simplified the process. To compute the normal approximation to the binomial distribution, take a simple random sample from a population. You must meet the conditions for a binomial distribution :

  • there are a certain number n of independent trials
  • the outcomes of any trial are success or failure
  • each trial has the same probability of a success p

Recall that if X is the binomial random variable, then X ~ B ( n, p ). The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np and nq must both be greater than five ( np >5 and nq >5; the approximation is better if they are both greater than or equal to 10). Then the binomial can be approximated by the normal distribution with mean μ = np and standard deviation σ = n p q . Remember that q = 1 – p . In order to get the best approximation, add 0.5 to x or subtract 0.5 from x (use x + 0.5 or x – 0.5). The number 0.5 is called the continuity correction factor and is used in the following example.

Suppose in a local Kindergarten through 12 th grade (K - 12) school district, 53 percent of the population favor a charter school for grades K through 5. A simple random sample of 300 is surveyed.

  1. Find the probability that at least 150 favor a charter school.
  2. Find the probability that at most 160 favor a charter school.
  3. Find the probability that more than 155 favor a charter school.
  4. Find the probability that fewer than 147 favor a charter school.
  5. Find the probability that exactly 175 favor a charter school.

Let X = the number that favor a charter school for grades K trough 5. X ~ B ( n, p ) where n = 300 and p = 0.53. Since np >5 and nq >5, use the normal approximation to the binomial. The formulas for the mean and standard deviation are μ = np and σ = n p q . The mean is 159 and the standard deviation is 8.6447. The random variable for the normal distribution is Y . Y ~ N (159, 8.6447). See The Normal Distribution for help with calculator instructions.

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Source:  OpenStax, Statistics i - math1020 - red river college - version 2015 revision a - draft 2015-10-24. OpenStax CNX. Oct 24, 2015 Download for free at http://legacy.cnx.org/content/col11891/1.8
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