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72% of the attendees at a basketball game support the home team. 50 attendees are randomly selected to receive a coupon for a free beverage from the concession stand. Use normal approximation to find the probability that 64% or less of those receiving the coupon support the home team. Round σ to four decimal places when performing calculations.

A) 10.93%
B) 10.75%
C) 10.56%
D) 10.38%



Answer :

To solve this problem using the normal approximation, we first need to calculate the mean and standard deviation of the distribution.

Given:
- Mean (μ) = 72%
- Probability of not supporting the home team = 100% - 72% = 28%
- Sample Size (n) = 50

Now, we can calculate the standard deviation (σ) using the formula:
σ = sqrt(np(1-p))

Substitute the values:
σ = sqrt(50 * 0.28 * 0.72)

Now, we can find the probability that 64% or less of those receiving the coupon support the home team by using the z-score formula:
z = (X - μ) / σ

Substitute X = 64%, μ = 72%, and σ from the previous calculation to find the z-score.

Finally, use a standard normal distribution table or calculator to find the probability corresponding to this z-score.

After performing the calculations, the probability that 64% or less of those receiving the coupon support the home team is approximately 10.93%. So, the answer is

A) 10.93%.

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