Question

After years of practicing at the local bowling alley, Allan has determined that his distribution of bowling scores is roughly symmetric, unimodal, and bell-shaped, with a mean of 182 points and a standard deviation of 23 points. How likely is it that Allan will roll a perfect game (300 points), just by random chance?

Answer 1

P(x = 300) = 1.45 × 10⁻⁷

Explanation:

This is a normal distribution problem with mean number of points = μ = 182 points

Standard deviation = σ = 23 points

Probability that Allan will roll a perfect game (300 points), just by random chance.

First of, we need to normalize/standardize 300.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (300 - 182)/23 = 5.13

300 is 5.13 Standard deviation from the mean

Probability of scoring 300 points = P(x = 300) = P(z = 5.13)

Using the normal distribution formula which is presented in the attached image to this question,

The mean = μ = 182

Standard deviation = σ = 23

x = variable whose probability is required = 300

P(x = 300) = P(z = 5.13) = 1.449193 × 10⁻⁷

Extremely unlikely!

Hope this helps!!!