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suppose that the distance of fly balls hit to the outfield (in baseball…

Question

suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 258 feet and a standard deviation of 53 feet. use your graphing calculator to answer the following questions. write your answers in percent form. round your answers to the nearest tenth of a percent. a) if one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled fewer than 204 feet? ( p(\text{fewer than } 204 \text{ feet})=) % b) if one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled more than 216 feet? ( p(\text{more than } 216 \text{ feet})=) % check answer question 13

Explanation:

Step1: Calculate z - score for part (a)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 258\), \(\sigma=53\), and \(x = 204\).

$$z=\frac{204 - 258}{53}=\frac{-54}{53}\approx - 1.02$$

Using a graphing calculator (normalcdf function: normalcdf\((-\infty,204,258,53)\) or normalcdf\((-1000,-1.02)\)), we find the probability.

Step2: Calculate z - score for part (b)

Using the z - score formula \(z=\frac{x-\mu}{\sigma}\), with \(\mu = 258\), \(\sigma = 53\), and \(x = 216\)

$$z=\frac{216-258}{53}=\frac{-42}{53}\approx - 0.79$$

Using the property \(P(X>x)=1 - P(X\leq x)\). So, \(P(X > 216)=1-\text{normalcdf}(-\infty,216,258,53)\) or \(1-\text{normalcdf}(-1000,-0.79)\)

Answer:

a) \(P(\text{fewer than }204\text{ feet})\approx15.4\%\)
b) \(P(\text{more than }216\text{ feet})\approx78.5\%\)