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:
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]
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.
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%.
