To solve this problem using the normal approximation, we will start by determining the parameters of the binomial distribution that describes the batting average.
In this case:
- The probability of success (getting a hit) is \( p = 0.320 \).
- The number of trials (at bats) is \( n = 100 \).
The number of hits \( X \) can be approximated by a binomial distribution, which can be approximated by a normal distribution when \( n \) is large.
Step 1: Calculate the mean and standard deviation of the binomial distribution.
The mean \( \mu \) of a binomial distribution is given by: \[ \mu = n \cdot p = 100 \cdot 0.320 = 32 \]
The standard deviation \( \sigma \) is given by: \[ \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{100 \cdot 0.320 \cdot (1 - 0.320)} \] Calculating \( 1 - p \): \[ 1 - p = 1 - 0.320 = 0.680 \] Now calculating \( \sigma \): \[ \sigma = \sqrt{100 \cdot 0.320 \cdot 0.680} = \sqrt{100 \cdot 0.2176} = \sqrt{21.76} \approx 4.655 \]
Step 2: Use the normal approximation to find the probability that the player gets at most 26 hits.
We want to find \( P(X \leq 26) \). With the continuity correction, we will look for \( P(X \leq 26.5) \).
We will standardize using the Z-score: \[ Z = \frac{X - \mu}{\sigma} = \frac{26.5 - 32}{4.655} \approx \frac{-5.5}{4.655} \approx -1.182 \]
Step 3: Look up the Z-score in a standard normal distribution table or use a calculator.
Using a standard normal distribution table or calculator to find \( P(Z \leq -1.182) \): \[ P(Z \leq -1.182) \approx 0.118 \]
So, rounding to the nearest thousandth, the probability that the player will get at most 26 hits at 100 at bats is approximately:
\[ \boxed{0.118} \]
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