Question

lebrons 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 220. (1 point)
76.27%
23.8%
15%
1.7%

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x - \mu}{\sigma}\), where \(x = 220\), \(\mu=180\), and \(\sigma = 28\).

$$z=\frac{220 - 180}{28}=\frac{40}{28}\approx1.43$$

Step2: Find the probability using the standard normal distribution

We want to find \(P(X>220)\), which is equivalent to \(P(Z > 1.43)\) in the standard normal distribution.
Since \(P(Z>z)=1 - P(Z\leq z)\), and using a standard normal table or technology (e.g., a calculator with a normalcdf function: normalcdf\((1.43,\infty)\) or in Excel: \(=1 - NORM.S.DIST(1.43,TRUE)\)), we find that \(P(Z\leq1.43)\approx0.9236\)

$$P(Z > 1.43)=1 - 0.9236 = 0.0764\approx7.6\%$$

Answer:

\(7.6\%\) (closest to \(7.7\%\) among the given options)