A job has a normal time of 12 minutes, a performance rating of 0.80, and an allowance of 20 percent of job time. The standard ti
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Answer:
a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.
b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use 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 question, we have that:
a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?
This is the pvalue of Z when X = 200 + 5 = 205 subtracted by the pvalue of Z when X = 200 - 5 = 195.
Due to the Central Limit Theorem, Z is:
X = 205
has a pvalue of 0.8413.
X = 195
has a pvalue of 0.1587.
0.8413 - 0.1587 = 0.6426
0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.
b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?
This is the pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 190.
X = 210
has a pvalue of 0.9772.
X = 195
has a pvalue of 0.0228.
0.9772 - 0.0228 = 0.9544
0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.