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Major League Baseball now records information about every pitch thrown in every game of every season. Statistician Jim Albert compiled data about every pitch thrown by 20 starting pitchers during the 2009 MLB season. The data set included the type of pitch thrown (curveball, changeup, slider, etc.) as well as the speed of the ball as it left the pitcher's hand. A histogram of speeds for all 30,740 fourseam fastballs thrown by these pitchers during the 2009 season is shown below, from which we can see that the speeds of these fastballs follow a Normal model with mean $\mu=92.12 \mathrm{mph}$ and a standard deviation of $\sigma=2.43 \mathrm{mph}$.
Compute the z-score of pitch with speed $92.8 \mathrm{mph}$. (Round your answer to two decimal places.)
Approximately what fraction of these four-seam fastballs would you expect to have speeds between $90.4 \mathrm{mph}$ and $92.8 \mathrm{mph}$ ? (Express your answer as a decimal, not a percent, and round to three decimal places.)
Approximately what fraction of these four-seam fastballs would you expect to have speeds above 92.8 $\mathrm{mph}$ ? (Express your answer as a decimal, not a percent, and round to three decimal places.)
A baseball fan wishes to identify the four-seam fastballs among the fastest $12 \%$ of all such pitches. Above what speed must a four-seam fastball be in order to be included in the fastest $12 \%$ ? (Round your answer to the nearest $0.1 \mathrm{mph}$.)
Answer
To find the speed corresponding to the fastest 12% of pitches, first find the z-score corresponding to 0.12 from a standard normal distribution table, which is approximately 1.175. Then use this z-score to find the corresponding pitch speed using the formula \(X = \mu + Z\sigma\). Substituting the given values, we get \(X = 92.12 + 1.175*2.43 = 94.99\) mph. Rounded to the nearest 0.1 mph, the speed must be \(\boxed{95.0}\) mph.
Step 1 ：Calculate the z-score for a pitch speed of 92.8 mph using the formula \(Z = \frac{X - \mu}{\sigma}\). Substituting the given values, we get \(Z = \frac{92.8 - 92.12}{2.43} = 0.28\).
Step 2 ：To find the fraction of pitches between 90.4 mph and 92.8 mph, first calculate the z-scores for these speeds. \(Z_{90.4} = \frac{90.4 - 92.12}{2.43} = -0.71\) and \(Z_{92.8} = \frac{92.8 - 92.12}{2.43} = 0.28\). The corresponding probabilities from a standard normal distribution table are 0.239 and 0.610 respectively. The fraction of pitches between these two speeds is the difference of these two probabilities, \(0.610 - 0.239 = 0.371\).
Step 3 ：To find the fraction of pitches above 92.8 mph, subtract the probability corresponding to the z-score of 92.8 mph from 1. This gives us \(1 - 0.610 = 0.390\).
Step 4 ：To find the speed corresponding to the fastest 12% of pitches, first find the z-score corresponding to 0.12 from a standard normal distribution table, which is approximately 1.175. Then use this z-score to find the corresponding pitch speed using the formula \(X = \mu + Z\sigma\). Substituting the given values, we get \(X = 92.12 + 1.175*2.43 = 94.99\) mph. Rounded to the nearest 0.1 mph, the speed must be \(\boxed{95.0}\) mph.
