Respuesta :
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 [tex]\mu[/tex] and standard deviation(which is the square root of the variance) [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex];
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:
[tex]\mu = 183, \sigma = \sqrt{121} = 11, n = 78, s = \frac{11}{\sqrt{78}} = 1.2455[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{184.7 - 183}{1.2455}[/tex]
[tex]Z = 1.365[/tex]
[tex]Z = 1.365[/tex] has a pvalue of 0.9139
X = 181.3
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{181.3 - 183}{1.2455}[/tex]
[tex]Z = -1.365[/tex]
[tex]Z = -1.365[/tex] 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
