Someone help me pleaseeeeee
2 answers:
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Answer:
0.9808 = 98.08% probability that the mean of a sample of 32 cars would differ from the population mean by less than 181 miles
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 can be 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
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
If he is correct, what is the probability that the mean of a sample of 32 cars would differ from the population mean by less than 181 miles?
This is the pvalue of Z when X = 4639 + 181 = 4820 subtracted by the pvalue of Z when X = 4639 - 181 = 4458. So
X = 4820
By the Central limit theorem
has a pvalue of 0.9904
X = 4458
has a pvalue of 0.0096
0.9904 - 0.0096 = 0.9808
0.9808 = 98.08% probability that the mean of a sample of 32 cars would differ from the population mean by less than 181 miles