Math Problem Statement

The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 8.9 cm. b. Find the probability that the mean for 20 randomly selected distances is greater than 203.20 cm c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

Solution

Problem Breakdown

Given Data:

  • Mean (μ\mu) = 205 cm
  • Standard deviation (σ\sigma) = 8.9 cm
  • Sample size (nn) = 20
  • Question asks for:
    1. (b): Probability that the sample mean is greater than 203.20 cm.
    2. (c): Explanation of why the normal distribution is applicable despite n<30n < 30.

(b) Finding Probability for Xˉ>203.20\bar{X} > 203.20:

  1. Step 1: Standard Error of the Mean (SEXˉ\text{SE}_{\bar{X}}): The standard error of the mean is calculated as:

    SEXˉ=σn\text{SE}_{\bar{X}} = \frac{\sigma}{\sqrt{n}}

    Substituting the values: SEXˉ=8.9201.99cm.\text{SE}_{\bar{X}} = \frac{8.9}{\sqrt{20}} \approx 1.99 \, \text{cm}.

  2. Step 2: Z-Score Calculation: The Z-score for the sample mean of 203.20 cm is given by:

    Z=XˉμSEXˉZ = \frac{\bar{X} - \mu}{\text{SE}_{\bar{X}}}

    Substituting the values: Z=203.202051.990.90.Z = \frac{203.20 - 205}{1.99} \approx -0.90.

  3. Step 3: Find the Probability: Using standard normal distribution tables or a calculator, the cumulative probability for Z=0.90Z = -0.90 is approximately 0.18410.1841.

    The probability that the sample mean is greater than 203.20 cm is: P(Xˉ>203.20)=10.1841=0.8159.P(\bar{X} > 203.20) = 1 - 0.1841 = 0.8159.

    Final Answer for (b): The probability is approximately 0.816 (81.6%).


(c) Why is the Normal Distribution Applicable?

The normal distribution can be used for the sampling distribution of the mean even when n<30n < 30 because:

  1. The underlying population is normally distributed.
  2. The Central Limit Theorem (CLT) states that for any sample size, the sampling distribution of the sample mean will be normal if the population itself is normal.

In this case, it is explicitly stated that the overhead reach distances of adult females are normally distributed, so the sample mean will also follow a normal distribution, regardless of the sample size being less than 30.


Summary:

  • (b): The probability P(Xˉ>203.20)0.816P(\bar{X} > 203.20) \approx 0.816 or 81.6%.
  • (c): The normal distribution is valid because the population is already normally distributed, ensuring that the sample mean follows a normal distribution regardless of sample size.

Do you want more detailed steps or a plot of the normal distribution?


Related Questions:

  1. What is the Z-score for Xˉ=207\bar{X} = 207?
  2. How does increasing the sample size affect the standard error of the mean?
  3. If n=50n = 50, what would P(Xˉ>203.20)P(\bar{X} > 203.20) be?
  4. How does the Central Limit Theorem apply when the population is not normal?
  5. What is the probability that the sample mean is less than 201 cm?

Tip: Always check whether the population distribution is normal or if the sample size is large enough for the Central Limit Theorem to apply.

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Standard Error
Z-Score
Central Limit Theorem

Formulas

Standard Error of the Mean (SE_x) = σ / √n
Z-Score Formula: Z = (X̄ - μ) / SE_x

Theorems

Central Limit Theorem

Suitable Grade Level

Grades 10-12