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?

P(x = 300) = 1.45 × 10⁻⁷

Step-by-step 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!!!

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step-by-step explanation:

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let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b