Round 72.5 to the nearest hundredth
1 answer:
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Answer:
61.70% approximate probability that X will be more than 0.5 away from the population mean
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
and standard deviation
In this problem, we have that:
What is the approximate probability that X will be more than 0.5 away from the population mean?
This is the probability that X is lower than 36-0.5 = 35.5 or higher than 36 + 0.5 = 36.5.
Lower than 35.5
Pvalue of Z when X = 35.5. So
By the Central Limit Theorem
has a pvalue of 0.3085.
30.85% probability that X is lower than 35.5.
Higher than 36.5
1 subtracted by the pvalue of Z when X = 36.5. SO
has a pvalue of 0.6915.
1 - 0.6915 = 0.3085
30.85% probability that X is higher than 36.5
Lower than 35.5 or higher than 36.5
2*30.85 = 61.70
61.70% approximate probability that X will be more than 0.5 away from the population mean