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2 answers:
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Answer:
83.89% probability that the sample mean would be greater than 98.8 gallons
Step-by-step explanation:
To solve this question, we have 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 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 \mu and standard deviation, which is also called standard error
In this problem, we have that:
The standard deviation is the square root of the variance. So
If 44 racing cars are randomly selected, what is the probability that the sample mean would be greater than 98.8 gallons
This probability is 1 subtracted by the pvalue of Z when X = 98.8. So
By the Central Limit Theorem
has a pvalue of 0.1611
1 - 0.1611 = 0.8389
83.89% probability that the sample mean would be greater than 98.8 gallons
The first part of the answer can be found by finding the GCF between 15 and 25. 15 is divisible by 1, 3, 5, and 15. 25 is divisible by 1, 5, and 25. You can see that their GCF is 5. Next, you need to look at the variables. The first monomial has a, b, and 
. The second monomial has 
, b, and c. They each have at least one a, one b, and one c, so those will be in the GCF (abc). If you put that together with the 5, you get that the GCF is 5abc.