One statistic used to assess professional golfers is driving accuracy, the percent of drives that land in the fairway.

Driving accuracy for PGA Tour professionals ranges from about 40% to about 75%.
Tiger Woods hits the fairway about 60% of the time.
One reason why the Normal approximation may fail to give accurate estimates of binomial probabilities is that the binomial distributions are discrete and the Normal distributions are continuous.
That is, counts take only whole number values but Normal variables can take any value.
We can improve the Normal approximation by treating each whole number count as if it occupied the interval from 0.5 below the number to 0.5 above the number.
For example, approximate a binomial probability P(X ¡Ý 10) by finding the Normal probability P(X ¡Ý 9.5).
Be careful: binomial P(X > 10) is approximated by Normal P(X ¡Ý 10.5).
We saw in Exercise 13.30 (above) that Tiger Woods hits the fairway in 60% of his drives.
We will assume that his drives are independent and that each has probability 0.6 of hitting the fairway.
Tiger drives 25 times.
The exact binomial probability that he hits 15 or more fairways is 0.5858.

What is the Normal approximation to P(X ¡Ý 15)?
Do not use the continuity correction.
A. 0.5
B. 0.68
C. 0.95
D. 0.75

What is the Normal approximation using the continuity correction? Use Table A to answer the question.
A. 0.4207
B. 0.1586
C. 0.5793
D. 0.6591

To find the Normal approximation to P(X ≥ 15) without using the continuity correction, we can calculate the mean (μ) and standard deviation (σ) of the binomial distribution and then use the Normal distribution to estimate the probability.

Given that Tiger Woods hits the fairway with a probability of 0.6 and drives 25 times, we can use the formula for the mean of a binomial distribution to find μ:
μ = n * p
μ = 25 * 0.6
μ = 15

Similarly, we can use the formula for the standard deviation of a binomial distribution to find σ:
σ = sqrt(n * p * (1 - p))
σ = sqrt(25 * 0.6 * (1 - 0.6))
σ = sqrt(6)

Now, we can use these values to approximate P(X ≥ 15) using the Normal distribution. Since we are not using the continuity correction, we will use P(X > 14.5) instead.
P(X > 14.5) can be written as P(Z > (14.5 - μ) / σ), where Z is a standard Normal random variable.

Substituting the values, we have:
P(Z > (14.5 - 15) / sqrt(6))
P(Z > -0.5 / sqrt(6))
P(Z > -0.2041)

Using a standard Normal distribution table or a statistical calculator, we can find the probability that Z is greater than -0.2041. This probability is approximately 0.5793.

Therefore, the Normal approximation to P(X ≥ 15) without using the continuity correction is approximately 0.5793.

The answer is:
C. 0.5793

To find the Normal approximation using the continuity correction, we need to use P(X ≥ 15.5):
P(X ≥ 15.5) can be written as P(Z ≥ (15.5 - μ) / σ).

Substituting the values, we have:
P(Z ≥ (15.5 - 15) / sqrt(6))
P(Z ≥ 0.5 / sqrt(6))
P(Z ≥ 0.2041)

Again, using a standard Normal distribution table or a statistical calculator, we can find the probability that Z is greater than or equal to 0.2041. This probability is approximately 0.4207.

Therefore, the Normal approximation to P(X ≥ 15) using the continuity correction is approximately 0.4207.

The answer is:
A. 0.4207