Answer:
0.8278 = 82.78% probability that the mean of the sample would differ from the population mean by less than 1.7 watts
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(which is the square root of the variance)
, 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:
Probability that the mean of the sample would differ from the population mean by less than 1.7 watts?
This is the pvalue of Z when X = 183+1.7 =184.7 subtracted by the pvalue of Z when X = 183 - 1.7 = 181.3.
X = 184.7
By the Central Limit Theorem
has a pvalue of 0.9139
X = 181.3
has a pvalue of 0.0861
0.9139 - 0.0861 = 0.8278
0.8278 = 82.78% probability that the mean of the sample would differ from the population mean by less than 1.7 watts