High School

Question 18: The weights of women in one age group are normally distributed with a standard deviation of 20 lb. A researcher wishes to estimate the mean weight of all women in this age group. Find how large a sample must be drawn in order to be 90% confident that the sample mean will not differ from the population mean by more than 3.5 lb.

Group of answer choices:
A. 86
B. 44
C. 126
D. 98
E. 89

Answer :

To solve this problem, we need to determine the sample size required for a given level of confidence and margin of error for estimating the mean weight of women in a certain age group.

Given:

  • Standard Deviation (C3): 20 lb
  • Confidence Level: 90%
  • Margin of Error (E): 3.5 lb

Step-by-Step Solution:

  1. Find the Z-Score for the Confidence Level:

    For a 90% confidence level, we look up the Z-score that corresponds to a cumulative probability of 0.95 (since 90% confidence means 5% of the distribution is split equally at both tails, leaving 95% in the middle). This Z-score is generally 1.645.

    [tex]Z = 1.645[/tex]

  2. Apply the Formula for Sample Size (n):

    The formula for determining the required sample size is:

    [tex]n = \left( \frac{Z 3C3}{E} \right)^2[/tex]

    Plug in the values:

    [tex]n = \left( \frac{1.645 0}{3.5} \right)^2[/tex]

  3. Calculate:

    First, calculate [tex]\frac{1.645 20}{3.5}[/tex]:

    [tex]\frac{1.645 20}{3.5} \approx 9.4[/tex]

    Then square the result:

    [tex]n = (9.4)^2 \approx 88.36[/tex]

    Since we can't have a fraction of a sample, round up to the nearest whole number to ensure that the sample is large enough:

    [tex]n = 89[/tex]

Therefore, the researcher must draw a sample of at least 89 women to be 90% confident that the sample mean will differ by no more than 3.5 lb from the population mean.

Conclusion:

The answer to the question is (E) 89. The researcher needs to draw a sample of 89 women to meet the desired confidence and margin of error.

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