Respuesta :
Answer:
a) 1.19% of the baseballs produced by this factory are too heavy for use by Major League Baseball
b) 94.97% of the baseballs produced by this factory are acceptable for use by Major League Baseball
c) 0.19% probability that the average weight of the balls the coach purchases is greater than 5.15 ounces
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
[tex]\mu = 5.11, \sigma = 0.062[/tex]
a. What proportion of the baseballs produced by this factory are too heavy for use by Major League Baseball?
Heavier than 5.25 ounces, which is 1 subtracted by the pvalue of Z when X = 5.25. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5.25 - 5.11}{0.062}[/tex]
[tex]Z = 2.26[/tex]
[tex]Z = 2.26[/tex] has a pvalue of 0.9881
1 - 0.9881 = 0.0119
1.19% of the baseballs produced by this factory are too heavy for use by Major League Baseball
b. What proportion of the baseballs produced by this factory are acceptable for use by Major League Baseball?
Between 5 and 5.25 ounces, which is the pvalue of Z when X = 5.25, subtracted by the pvalue of Z when X = 5. So
X = 5.25
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5.25 - 5.11}{0.062}[/tex]
[tex]Z = 2.26[/tex]
[tex]Z = 2.26[/tex] has a pvalue of 0.9881
X = 5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5 - 5.11}{0.062}[/tex]
[tex]Z = -1.77[/tex]
[tex]Z = -1.77[/tex] has a pvalue of 0.0384
0.9881 - 0.0384 = 0.9497
94.97% of the baseballs produced by this factory are acceptable for use by Major League Baseball
c. A coach purchases 20 baseballs from this factory. What is the probability that the average weight of the balls the coach purchases is greater than 5.15 ounces?
Now we have [tex]n = 20, s = \frac{0.062}{\sqrt{20}} = 0.0139[/tex]
This probability is 1 subtracted by the pvalue of Z when X = 5.15. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{5.15 - 5.11}{0.0139}[/tex]
[tex]Z = 2.89[/tex]
[tex]Z = 2.89[/tex] has a pvalue of 0.9981
1 - 0.9981 = 0.0019
0.19% probability that the average weight of the balls the coach purchases is greater than 5.15 ounces
