LeBron’s average bowling score for the season is 180 with a standard deviation of 28. Use technology to determine which of the following represents the probability that LeBron records a score higher than 200.(1 point) Responses 17% 17% 76.2% 76.2% 1.1% 1.1% 23.8%

1 answer

To find the probability that LeBron records a score higher than 200, we'll use the properties of the normal distribution since we know the mean and standard deviation.

  1. Identify the parameters:

    • Mean (\( \mu \)) = 180
    • Standard Deviation (\( \sigma \)) = 28
  2. Calculate the z-score for 200: \[ z = \frac{X - \mu}{\sigma} = \frac{200 - 180}{28} = \frac{20}{28} \approx 0.7143 \]

  3. Use the z-score to find the probability: You can look up the z-score in a standard normal distribution table or use a calculator to find the probability of z being less than 0.7143.

    Using a standard normal distribution table or calculator, the probability \( P(Z < 0.7143) \) approximately equals 0.7611 or 76.11%. Therefore, the probability of scoring higher than 200 is: \[ P(X > 200) = 1 - P(Z < 0.7143) \approx 1 - 0.7611 = 0.2389 \text{ or } 23.89% \]

From the provided options, the probability that LeBron records a score higher than 200 is closest to 23.8%.