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Chapter 8: Problem 32

The mean batting average in major league baseball is about \(0.250\). Supposing that batting averages are normally distributed, that the standard deviation in the averages is \(0.05\), and that there are 250 batters, what is the expected number of batters with an average of at least \(0.400 ?^{54}\)

Short Answer

Expert verified
The expected number of batters with an average of at least \(0.400\) is \(0\).

Step by step solution

01

Calculate the z-score

First, we need to calculate the z-score for a batting average of \(0.400\). The z-score represents the number of standard deviations away from the mean (\(0.250\)) the batting average is. To calculate the z-score, we can use the formula: $$ z = \frac{x - \mu}{\sigma} $$ where \(x\) is the value of interest (\(0.400\)), \(\mu\) is the mean (\(0.250\)), and \(\sigma\) is the standard deviation (\(0.05\)). So, we have: $$ z = \frac{0.400 - 0.250}{0.05} $$
02

Evaluate the z-score

Now, let's evaluate the z-score: $$ z = \frac{0.400 - 0.250}{0.05} = \frac{0.150}{0.05} = 3 $$ Therefore, the z-score for a batting average of \(0.400\) is \(3\). It means that batting average is 3 standard deviations away from the mean.
03

Calculate the probability

Next, we need to calculate the probability of having a batting average of at least \(0.400\). We can calculate this using the complement rule: $$ P(x \geq 0.400) = 1 - P(x < 0.400) $$ We can find \(P(x < 0.400)\) using the z-score (\(3\)) and the normal distribution table which shows the area to the left of the z-score. Looking up the z-score of \(3\) in the table, we find that the area to the left is approximately \(0.9987\). So, the probability of having a batting average of at least \(0.400\) is: $$ P(x \geq 0.400) = 1 - 0.9987 = 0.0013 $$
04

Calculate the expected number of batters

Finally, we can calculate the expected number of batters with an average of at least \(0.400\). To do this, we can multiply the probability we found in the last step by the total number of batters (\(250\)): $$ Expected\,Number\,of\,Batters = Probability \times Number\,of\,Batters $$ $$ Expected\,Number\,of\,Batters = 0.0013 \times 250 $$
05

Evaluate the expected number of batters

Now, let's evaluate the expected number of batters: $$ Expected\,Number\,of\,Batters = 0.0013 \times 250 = 0.325 $$ Since we can't have a fraction of a batter, we round the result to the nearest whole number. Therefore, the expected number of batters with an average of at least \(0.400\) is \(0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Z-Score Calculation
Understanding the z-score is crucial when dealing with normally distributed data sets. The z-score, also known as the standard score, indicates how many standard deviations an element is from the mean. In the realm of baseball averages, for example, calculating a z-score lets us quantify a player's performance relative to all other players.

To calculate the z-score, you apply the formula:
\[\begin{equation} z = \frac{x - \mu}{\sigma}\end{equation}\]
where the batting average you're interested in is represented by \(x\), the mean average by \(\mu\), and the standard deviation by \(\sigma\). If a player's average is significantly higher than the mean, they'll have a high z-score, which in this context would mean a standout performance.
Probability Theory
Probability theory is the mathematical framework that underpins the study of random phenomena. When we speak of the 'probability' of a baseball player having a certain batting average, we're asking how likely it is that this player's average falls within a specific range on the normal distribution curve.

Probability values range from 0 to 1, where 0 means the event is impossible and 1 means it's certain. To find the likelihood of a player having an average above a certain threshold, we can use the complement rule: \[\begin{equation} P(x \geq a) = 1 - P(x < a)\end{equation}\]
This allows us to use standard z-score tables or software to find the probability of averages less than the threshold, and then deduct it from 1 to find the probability of the opposite (averages being at least that threshold).
Standard Deviation
The concept of standard deviation is essential for describing the variability within a data set. Standard deviation measures the amount of variation or dispersion of a set of values. A low standard deviation means that most of the numbers are close to the mean (average) of the set. A high standard deviation indicates that the numbers are spread out over a wider range.

In the context of our baseball example, a standard deviation of \(0.05\) means that the majority of batting averages lie within 0.05 of the mean on either side, hence most players' averages cluster around 0.250. Batters with averages far from 0.250 are more exceptional, as indicated by high z-scores.

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Most popular questions from this chapter

Compute the (sample) variance and standard deviation of the given data sample. (You calculated the means in the Section 8.3 exercises. Round all answers to two decimal places.) 3,1,6,-3,0,5

Your company, Sonic Video, Inc., has conducted research that shows the following probability distribution, where \(X\) is the number of video arcades in a randomly chosen city with more than 500,000 inhabitants. $$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \boldsymbol{x} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x}) & .07 & .09 & .35 & .25 & .15 & .03 & .02 & .02 & .01 & .01 \\ \hline \end{array} $$ a. Compute \(\mu=E(X)\) and interpret the result. HINT [See Example 3.] b. Which is larger, \(P(X<\mu)\) or \(P(X>\mu) ?\) Interpret the result.

Your pet tarantula, Spider, has a .12 probability of biting an acquaintance who comes into contact with him. Next week, you will be entertaining 20 friends (all of whom will come into contact with Spider). a. How many guests should you expect Spider to bite? b. At your last party, Spider bit 6 of your guests. Assuming that Spider bit the expected number of guests, how many guests did you have?

The new computer your business bought lists a mean time between failures of 1 year, with a standard deviation of 2 months. Ten months after a repair, it breaks down again. Is this surprising? (Assume that the times between failures are normally distributed.)

A roulette wheel has the numbers 1 through 36,0 , and \(00 .\) A bet on two numbers pays 17 to 1 (that is, if one of the two numbers you bet comes up, you get back your \(\$ 1\) plus another \(\$ 17\) ). How much do you expect to win with a \(\$ 1\) bet on two numbers? HINT [See Example 4.]
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