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The normal distribution for women's height in North America has \(\mu=65\) inches, \(\sigma=3.5\) inches. Most major airlines have height requirements for flight attendants (www.cabincrewjobs.com). Although exceptions are made, the minimum height requirement is 62 inches. What proportion of adult females in North America are not tall enough to be a flight attendant?

Short Answer

Expert verified
Approximately 19.57% of adult females in North America are not tall enough to be a flight attendant.

Step by step solution

01

Understand the Problem

We are given a normal distribution for women's height with \(\mu = 65\) inches and \(\sigma = 3.5\) inches. The problem asks us to find the proportion of women whose height is less than 62 inches.
02

Convert Height to a Z-score

To find the proportion, we first convert the height requirement to a Z-score using the formula: \[ Z = \frac{X - \mu}{\sigma} \]where \(X = 62\), \(\mu = 65\), and \(\sigma = 3.5\). \[ Z = \frac{62 - 65}{3.5} = \frac{-3}{3.5} \approx -0.857 \]
03

Find the Proportion from Z-table

Now, we need to find the proportion or probability of a Z-score being less than \(-0.857\) using a standard normal distribution table (Z-table). This gives the cumulative probability of a score being less than \(-0.857\).
04

Interpret the Z-table Value

Look up the Z-score of \(-0.857\) in the Z-table to find the cumulative probability. Typically, this value will be approximately 0.1957, indicating that about 19.57% of women are shorter than 62 inches.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Women's Height
When we talk about women's height in North America, we're dealing with a set of data that often forms a normal distribution. This is a common way of representing data in statistics because it showcases how most women will be around an average height, with fewer women being significantly taller or shorter.
The average height, or mean, in this case, is 65 inches, and the standard deviation is 3.5 inches.
Understanding these two measures is essential because they help us comprehend how much variation there is from the average height.
  • The average, \( \mu \), is the typical height that you would expect.
  • The standard deviation, \( \sigma \), provides insights into the spread of heights.
    A small \( \sigma \) means most heights are clustered around the average, while a larger \( \sigma \) means more variation.
This normal distribution becomes a crucial factor when determining how many women meet specific criteria, like a height requirement for certain jobs.
Z-score Calculation
To find out how many women fall below the minimum height requirement, we use something known as a Z-score. A Z-score tells us how many standard deviations away a particular value is from the mean. This is helpful for assessing where a certain measurement falls in a distribution.
To calculate the Z-score, we use the formula: \[ Z = \frac{X - \mu}{\sigma} \], where:
  • \( X = 62 \) inches (minimum height requirement).
  • \( \mu = 65 \) inches (average height).
  • \( \sigma = 3.5 \) inches (standard deviation).
Plugging the numbers into the formula gives us a Z-score of approximately \( -0.857 \). This implies that 62 inches is \( 0.857 \) standard deviations below the mean, indicating how this particular height compares to the average.
Standard Normal Distribution
The concept of a standard normal distribution is fundamental to statistics and useful in many real-life applications. In this context, the standard normal distribution is a special normal distribution with a mean of 0 and a standard deviation of 1. Instead of measuring absolute values like height, it measures Z-scores, which reflect how data compares to the average. This allows us to compare diverse datasets on a standard scale.By standardizing the data to a Z-score, we can refer to a Z-table to find probabilities and proportions related to any Z-score value.
  • The Z-table helps translate the Z-score back to an understandable proportion of data.
  • For instance, in our context, the Z-score of \( -0.857 \) is used to determine the proportion of women who are shorter than the minimum height requirement.
Utilizing the standard normal distribution simplifies the process of understanding where a particular measurement stands within a population.
Cumulative Probability
Cumulative probability is a crucial concept when dealing with normal distributions and Z-scores. It refers to the probability that a random variable is less than or equal to a certain value. This concept is directly applicable when figuring out how many women do not meet a specific height threshold.
When you calculate a Z-score, as we did earlier, you can then reference a Z-table to find the cumulative probability. The Z-table shows the probability of a given Z-score or less. In our scenario, a Z-score of \(-0.857\) corresponds to a cumulative probability of approximately 0.1957.
  • This means about 19.57% of women are below the height requirement.
  • A cumulative probability near this number suggests that only a small portion of females fall short of the height needed to be a flight attendant.
  • It also indicates how commonly such values occur, helping organizations set fair thresholds.
Understanding cumulative probability aids significantly in making informed decisions and predictions based on statistical data.

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Most popular questions from this chapter

Jane Doe claims to possess extrasensory perception (ESP). She says she can guess more often than not the outcome of a flip of a balanced coin in another room. In an experiment, a coin is flipped three times. If she does not actually have ESP, find the probability distribution of the number of her correct guesses. a. Do this by constructing a sample space, finding the probability for each point, and using them to construct the probability distribution. b. Do this using the formula for the binomial distribution.

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The state of Ohio has several statewide lottery options. One is the Pick 3 game in which you pick one of the 1000 three-digit numbers between 000 and 999\. The lottery selects a three-digit number at random. With a bet of \(\$ 1,\) you win \(\$ 500\) if your number is selected and nothing (\$0) otherwise. (Many states have a very similar type of lottery.) (Source: Background information from www.ohiolottery.com.) a. With a single \(\$ 1\) bet, what is the probability that you win \(\$ 500 ?\) b. Let \(X\) denote your winnings for a \(\$ 1\) bet, so \(x=\$ 0\) or \(x=\$ 500\). Construct the probability distribution for \(X\). c. Show that the mean of the distribution equals 0.50 , corresponding to an expected return of 50 cents for the dollar paid to play. Interpret the mean. d. In Ohio's Pick 4 lottery, you pick one of the 10,000 four-digit numbers between 0000 and 9999 and (with a \(\$ 1\) bet ) win \(\$ 5000\) if you get it correct. In terms of your expected winnings, with which game are you better off - playing Pick 4 , or playing Pick 3 in which you win \(\$ 500\) for a correct choice of a three-digit number? Justify your answer.

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