a) Probability that a randomly selected ball is within regulation weight = 0.9032
b) Probability that at least 15 of the 16 baseballs in a pack are within regulation weight = 0.5324
c) Probability that a pack of 16 baseballs would have an average weight of more than 147 grams = 0.04093
Step-by-step explanation:
Complete Question
A regulation baseball can weigh no more than 149 grams. A factory produces baseballs with weights that are normally distributed with a mean of 146 grams and a standard deviation of 2.3 grams.
(a) If a baseball produced by the factory is randomly selected, what is the probability that it is within regulation weight?
(b) The baseballs are shipped in boxes of 16. What is the probability that at least 15 of the 16 baseballs in a pack are within regulation weight?
(c) The factory will not ship a box of 16 if the average weight of the baseballs in the box exceeds 147 grams. What is the probability that a pack of 16 baseballs would have an average weight of more than 147 grams?
Solution
a) This is a normal distribution problem with
Mean = μ = 146 g
Standard deviation = σ = 2.3 g
The baseballs fron the factory cannot weigh more than 149 g, so, we can now calculate the proportion of the balls in the factory that weigh within the regulation range
Within the regulation limit is P(x < 149)
We first normalize or standardize 149
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (149 - 146)/2.6 = 1.30
To determine the required probability
P(x < 149) = P(z < 1.30)
We'll use data from the normal distribution table for these probabilities
P(x < 149) = P(z < 1.30) = 0.9032
b) Probability that a ball made in the factory is within regulation = 0.9032
We now want the probability that at least 15 balls out of 16 are within regulation.
This becomes a binomial distribution problem
A binomial experiment is one in which the probability of success doesn't change with every run or number of trials (each ball had an equal probability of being within the regulation limit). It usually consists of a number of runs/trials with only two possible outcomes, a success or a failure (16 balls, it is either they are within regulation or not).
The outcome of each trial/run of a binomial experiment is independent of one another.
Probability that at least 15 balls out of 16 are within regulation = P(X ≥ 15)
P(X ≥ 15) = P(X = 15) + P(X = 16)
Binomial distribution function is represented by
P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ
n = total number of sample spaces = 16 balls
x = Number of successes required = at least 15, so, 15, 16.
p = probability of success = probability that a ball is within the regulation limit = 0.9032
q = probability of failure = probability that a ball is NOT within the regulation limit = 1 - p = 1 - 0.9032 = 0.0968
P(X = 15) = ¹⁶C₁₅ (0.9032)¹⁵ (0.0968)¹⁶⁻¹⁵ = 0.33632122307 = 0.3363
P(X = 16) = ¹⁶C₁₆ (0.9032)¹⁶ (0.0968)¹⁶⁻¹⁶ = 0.19612947358 = 0.1961
P(X ≥ 15) = P(X = 15) + P(X = 16) = 0.3363 + 0.1961 = 0.5324
c) This becomes working with a sampling distribution of sample size 16.
The mean of sampling distribution is the same as the population mean.
μₓ = μ = 146 g
But the standard deviation of the sampling distribution is given as
σₓ = (σ/√N)
σ = population standard deviation = 2.3 g
N = Sample size = 16
σₓ = (2.3/√16) = 0.575 g
Probability that a pack of 16 baseballs would have an average weight of more than 147 grams = P(x > 147)
We normalize or standardize 147
z = (x - μₓ)/σₓ = (147 - 146)/0.575 = 1.74
The required probability
P(x > 147) = P(z > 1.74)
We'll use data from the normal distribution table for these probabilities
P(x > 147) = P(z > 1.74) = 1 - P(z ≤ 1.74)
= 1 - 0.95907
= 0.04093
Hope this Helps!!!
