Respuesta :
Answer:
33.20% probability that the sample mean would differ from the population mean by greater than 1 gallon
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 [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 = 114, \sigma = 7, n = 46, s = \frac{7}{\sqrt{46}} = 1.03[/tex]
Either the sample mean differs by 1 gallon or less, or it differs by more than 1 gallon. The sum of the probabilities of these events is decimal 1.
Probability it differs by 1 gallon or less
pvalue of Z when X = 114+1 = 115 subtracted by the pvalue of Z when X = 114-1 = 113.
X = 115
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{115 - 114}{1.03}[/tex]
[tex]Z = 0.97[/tex]
[tex]Z = 0.97[/tex] has a pvalue of 0.8340
X = 113
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{113 - 114}{1.03}[/tex]
[tex]Z = -0.97[/tex]
[tex]Z = -0.97[/tex] has a pvalue of 0.1660
0.8340 - 0.1660 = 0.6680
0.6680 probability it differs from the population mean by 1 gallon or less.
Probability it differs from the population mean by greater than 1 gallon.
0.6680 + p = 1
p = 0.3320
33.20% probability that the sample mean would differ from the population mean by greater than 1 gallon
