Robert is purchasing some packs of chicken from the supermarket. The first pack weights 45 ounces . The second pack weights 3 po
2 answers:
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Answer:
a) 0.259
b) 0.297
c) 0.497
Step-by-step explanation:
To solve this problem, it is important to know 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 and standard deviation
, the zscore of a measure X is given by:
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 random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?
Here we have that
This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.
X = 11.2
By the Central Limit Theorem
has a pvalue of 0.6293.
X = 10.8
has a pvalue of 0.3707.
0.6293 - 0.3707 = 0.2586
0.259 probability, rounded to three decimal places.
b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.5 and 11 minutes?
Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So
X = 11
has a pvalue of 0.5.
X = 10.5
has a pvalue of 0.2033.
0.5 - 0.2033 = 0.2967
0.297, rounded to three decimal places.
c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?
Here we have that
This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.
X = 11.2
By the Central Limit Theorem
has a pvalue of 0.7486.
X = 10.8
has a pvalue of 0.2514.
0.7486 - 0.2514 = 0.4972
0.497, rounded to three decimal places.