A basketball player has a 76% chance of making a free throw. What is the probability of her making 104 or more free throws in 119 trials? Round your standard normal variable to two decimal places before using the table of values.
Use the normal approximation to the binomial distribution.
Your values are the following:
p = .76, q = 1 - p = .24, x = 104, and n = 119
Find mean and standard deviation.
mean = np = (119)(.76) = ?
sd = √npq = √(119)(.76)(.24) = ?
Once you calculate the mean and standard deviation, find z-score:
z = (x - mean)/sd
Once you have the z-score, check the z-table for the probability. Remember the question is asking the probability of 104 or more.
I hope this will help get you started.
To find the probability of making 104 or more free throws in 119 trials, we can use the normal distribution and the properties of the standard normal variable.
Step 1: Calculate the mean (μ) and standard deviation (σ) for the binomial distribution.
μ = n * p
= 119 * 0.76
= 90.44
σ = sqrt(n * p * (1 - p))
= sqrt(119 * 0.76 * (1 - 0.76))
= sqrt(21.9884)
≈ 4.69
Step 2: Standardize the value "104" using the standard normal distribution formula.
z = (x - μ) / σ
= (104 - 90.44) / 4.69
≈ 2.89
Step 3: Use the standard normal distribution table or calculator to find the probability associated with the standardized value.
Looking up the z-score of 2.89 in the standard normal distribution table, we find the corresponding probability to be approximately 0.9981.
Therefore, the probability of making 104 or more free throws in 119 trials is approximately 0.9981 or 99.81% (rounded to two decimal places).