Math Problem Statement

In a survey of a group of men, the heights in the 20-29 age group were normally distributed, with a mean of 68.3 inches and a standard deviation of 2.0 inches. A study participant is randomly selected. Complete parts (a) through (d) below. (a) Find the probability that a study participant has a height that is less than 66 inches. The probability that the study participant selected at random is less than 66 inches tall is

  • (Round to four decimal places as needed.) (b) Find the probability that a study participant has a height that is between 66 and 72 inches. The probability that the study participant selected at random is between 66 and 72 inches tall is (Round to four decimal places as needed.) (c) Find the probability that a study participant has a height that is more than 72 inches. The probability that the study participant selected at random is more than 72 inches tall is . (Round to four decimal places as needed.) (d) Identify any unusual events. Explain your reasoning. Choose the correct answer below. • A. The event in part (c) is unusual because its probability is less than 0.05. • B. The events in parts (a) and (b) are unusual because its probabilities are less than 0.05. • C. None of the events are unusual because all the probabilities are greater than 0.05. • D. The event in part (a) is unusual because its probability is less than 0.05.

Solution

Let's solve each part step-by-step.

Given Information

  • Mean (μ\mu) = 68.3 inches
  • Standard deviation (σ\sigma) = 2.0 inches

Part (a): Probability that a participant has a height less than 66 inches

  1. Calculate the z-score: z=Xμσ=6668.32=1.15z = \frac{X - \mu}{\sigma} = \frac{66 - 68.3}{2} = -1.15
  2. Find the probability using the standard normal distribution table or a calculator: P(Z<1.15)=0.1251P(Z < -1.15) = 0.1251 So, the probability that the study participant selected at random is less than 66 inches tall is 0.1251.

Part (b): Probability that a participant has a height between 66 and 72 inches

  1. Calculate the z-scores:
    • For 66 inches: z=6668.32=1.15z = \frac{66 - 68.3}{2} = -1.15
    • For 72 inches: z=7268.32=1.85z = \frac{72 - 68.3}{2} = 1.85
  2. Find the probability:
    • P(Z<1.15)=0.1251P(Z < -1.15) = 0.1251
    • P(Z<1.85)=0.9678P(Z < 1.85) = 0.9678
    • Probability between 66 and 72 inches: P(1.15<Z<1.85)=P(Z<1.85)P(Z<1.15)=0.96780.1251=0.8427P(-1.15 < Z < 1.85) = P(Z < 1.85) - P(Z < -1.15) = 0.9678 - 0.1251 = 0.8427 The probability that the participant's height is between 66 and 72 inches is 0.8427.

Part (c): Probability that a participant has a height more than 72 inches

  1. Calculate the z-score: z=7268.32=1.85z = \frac{72 - 68.3}{2} = 1.85
  2. Find the probability: P(Z>1.85)=1P(Z<1.85)=10.9678=0.0322P(Z > 1.85) = 1 - P(Z < 1.85) = 1 - 0.9678 = 0.0322 So, the probability that the study participant selected at random is more than 72 inches tall is 0.0322.

Part (d): Identify any unusual events

An event is generally considered unusual if its probability is less than 0.05.

  • In part (c), the probability is 0.0322, which is less than 0.05, making it an unusual event.

The correct answer is:

  • A. The event in part (c) is unusual because its probability is less than 0.05.

Let me know if you want further details on any part or have questions about any step.

Related Questions

  1. How would the probabilities change if the standard deviation were larger, say 3 inches instead of 2 inches?
  2. What would be the probability of selecting a participant with a height exactly equal to 68.3 inches?
  3. How do you interpret a probability of 0.8427 in the context of heights between 66 and 72 inches?
  4. What height marks the upper 5% of this distribution?
  5. How would these probabilities change if the mean height increased to 70 inches?

Tip

In a normal distribution, about 68% of values lie within one standard deviation from the mean, about 95% within two, and about 99.7% within three. This rule of thumb helps in estimating probabilities for normally distributed data.

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Math Problem Analysis

Mathematical Concepts

Probability
Normal Distribution
Standard Scores (Z-scores)

Formulas

Z-score formula: z = (X - μ) / σ
Standard normal distribution table for probability lookup

Theorems

68-95-99.7 Rule for normal distribution

Suitable Grade Level

College Statistics or High School AP Statistics