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Answer :
Answer:
1) 225 pounds
2) 4.6 pounds
3) 1.50% Ybar will be greater than 235 pounds
4) 33.36% probability that the weight of a single randomly selected college football player will be greater than 235 pounds
Step-by-step explanation:
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 = 225, \sigma = 23, n = 25, s = \frac{25}{\sqrt{25}} = 4.6[/tex]
1. the mean of Ybar
By the Central Limit Theorem, 225 pounds
2. the standard deviation of Ybar
By the Central Limit Theorem, 4.6 pounds
3. the probability that Ybar will be greater than 235 pounds
This is 1 subtracted by the pvalue of Z when X = 235. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{235 - 225}{4.6}[/tex]
[tex]Z = 2.17[/tex]
[tex]Z = 2.17[/tex] has a pvalue of 0.9850
1 - 0.9850 = 0.0150
1.50% Ybar will be greater than 235 pounds
4. the probability that the weight of a single randomly selected college football player will be greater than 235 pounds
This is 1 subtracted by the pvalue of Z when X = 235. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{235 - 225}{23}[/tex]
[tex]Z = 0.43[/tex]
[tex]Z = 0.43[/tex] has a pvalue of 0.6664
1 - 0.6664 = 0.3336
33.36% probability that the weight of a single randomly selected college football player will be greater than 235 pounds
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