Answer :
To solve this problem, it's important to understand that it involves the normal distribution, a continuous probability distribution that is symmetric around the mean. For a normal distribution, about 68% of data falls within one standard deviation (σ), 95% within two standard deviations, and 99.7% within three standard deviations from the mean ([tex]\mu[/tex]).
Given:
- Mean weight ([tex]\mu[/tex]): 151 lb
- Standard deviation ([tex]\sigma[/tex]): 15 lb
- Total students: 500
a. How many students weigh between 120 and 155 lb?
First, we calculate the z-scores for 120 lb and 155 lb:
[tex]\text{Z-score} = \frac{X - \mu}{\sigma}[/tex]
For 120 lb:
[tex]Z = \frac{120 - 151}{15} = \frac{-31}{15} \approx -2.07[/tex]For 155 lb:
[tex]Z = \frac{155 - 151}{15} = \frac{4}{15} \approx 0.27[/tex]
Now, we look up these z-scores in the standard normal distribution table (or use a calculator).
- The probability for [tex]Z = -2.07[/tex] is approximately 0.0192.
- The probability for [tex]Z = 0.27[/tex] is approximately 0.6064.
To find the probability of a student weighing between 120 and 155 lb, subtract the smaller probability from the larger one:
[tex]0.6064 - 0.0192 = 0.5872[/tex]
Now, calculate how many students this probability represents:
[tex]0.5872 \times 500 \approx 294[/tex]
So, approximately 294 students weigh between 120 and 155 lb.
b. What is the probability that a randomly selected male student weighs less than 128 pounds?
Calculate the z-score for 128 lb:
[tex]Z = \frac{128 - 151}{15} = \frac{-23}{15} \approx -1.53[/tex]
Look up [tex]Z = -1.53[/tex] in the standard normal distribution table (or use a calculator), which gives a probability of approximately 0.0630.
This means that there is a 6.30% probability that a randomly selected male student weighs less than 128 pounds.
