High School

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------------------------------------------------ Assume that weights are normally distributed with a mean of [tex]\mu = 143 \, \text{lb}[/tex] and a standard deviation of [tex]\sigma = 29 \, \text{lb}[/tex]. If 1 woman is randomly selected, find the probability that her weight is between 113 lb and 174 lb.

Answer :

To find the probability that a randomly selected woman's weight is between 113 lb and 174 lb, we can use the properties of the normal distribution.

The problem gives us:

  • Mean, [tex]\mu = 143[/tex] lb
  • Standard deviation, [tex]\sigma = 29[/tex] lb

We are looking for the probability that a woman's weight is between 113 lb and 174 lb.

Step-by-step Solution:

  1. Convert the weights to z-scores.

    The z-score is a way to describe a specific value's position relative to the mean in terms of standard deviations. It is calculated using the formula:

    [tex]z = \frac{x - \mu}{\sigma}[/tex]

    • For the lower end, 113 lb:
      [tex]z_1 = \frac{113 - 143}{29} = \frac{-30}{29} \approx -1.0345[/tex]

    • For the upper end, 174 lb:
      [tex]z_2 = \frac{174 - 143}{29} = \frac{31}{29} \approx 1.0689[/tex]

  2. Find the probability for each z-score.

    Using a standard normal distribution table or a calculator, we find:

    • The probability corresponding to [tex]z_1 \approx -1.0345[/tex] is approximately 0.150, meaning there is a 15% chance the weight is less than 113 lb.

    • The probability corresponding to [tex]z_2 \approx 1.0689[/tex] is approximately 0.857, meaning there is an 85.7% chance the weight is less than 174 lb.

  3. Calculate the probability between the two z-scores.

    To find the probability that a randomly selected woman's weight is between 113 lb and 174 lb, we subtract the cumulative probability up to 113 lb from the cumulative probability up to 174 lb:

    [tex]\text{Probability} = 0.857 - 0.150 = 0.707[/tex]

Final Answer:

The probability that a randomly selected woman weighs between 113 lb and 174 lb is approximately 70.7%.