Answer :
Answer: 0.5696 or 56.96%.
Step-by-step explanation:
To find the probability that the average number of head hits per player on a randomly selected team of 48 players is between 340 and 360, we can use the Central Limit Theorem (CLT). The CLT states that the distribution of the sample mean will be approximately normal if the sample size is large enough.
Given:
- Population mean ([tex]\(\mu\))[/tex] = 355 hits
- Population standard deviation [tex](\(\sigma\))[/tex] = 80 hits
- Sample size [tex](\(n\))[/tex] = 48 players
First, we need to calculate the standard error of the mean (SEM):
[tex]\[ \text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{80}{\sqrt{48}} \approx 11.547 \][/tex]
Next, we standardize the values 340 and 360 to find the corresponding z-scores.
For [tex]\( X = 340 \)[/tex]:
[tex]\[ z = \frac{X - \mu}{\text{SEM}} = \frac{340 - 355}{11.547} \approx -1.30 \][/tex]
For [tex]\( X = 360 \):[/tex]
[tex]\[ z = \frac{X - \mu}{\text{SEM}} = \frac{360 - 355}{11.547} \approx 0.43 \][/tex]
Now, we look up these z-scores in the standard normal distribution table or use a calculator to find the probabilities.
For [tex]\( z = -1.30 \)[/tex]:
[tex]\[ P(Z < -1.30) \approx 0.0968 \][/tex]
For [tex]\( z = 0.43 \)[/tex]:
[tex]\[ P(Z < 0.43) \approx 0.6664 \][/tex]
To find the probability that the average number of head hits per player is between 340 and 360, we subtract the smaller probability from the larger one:
[tex]\[ P(340 < \bar{X} < 360) = P(Z < 0.43) - P(Z < -1.30) \approx 0.6664 - 0.0968 = 0.5696 \][/tex]
So, the probability that the average number of head hits per player is between 340 and 360 is approximately 0.5696 or 56.96%.
