A sample of 11 students using a Ti-89 calculator averaged 31.5 items identified, with a standard deviation of 8.35. A sample of
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Answer:
64.76% probability that the mean diameter of the sample shafts would differ from the population mean by less than .2 inches
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 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 normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
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:
What is the probability that the mean diameter of the sample shafts would differ from the population mean by less than .2 inches?
This is the pvalue of Z when X = 200 + 0.2 = 200.2 subtracted by the pvalue of Z when X = 200 - 0.2 = 199.8. So
X = 200.2
By the Central Limit Theorem
has a pvalue of 0.8238
X = 199.8
has a pvalue of 0.1762
0.8238 - 0.1762 = 0.6476
64.76% probability that the mean diameter of the sample shafts would differ from the population mean by less than .2 inches