Respuesta :
Answer:
0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months
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 can be 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 random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means with size n of at least 30 can be approximated to a normal distribution with mean
In this problem, we have that:
[tex]\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.6275[/tex]
What is the probability that the sample mean would differ from the true mean by less than 1.11 months?
This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So
X = 120.1
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{120.1 - 119}{1.6275}[/tex]
[tex]Z = 0.68[/tex]
[tex]Z = 0.68[/tex] has a pvalue of 0.7517
X = 117.9
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{117.9 - 119}{1.6275}[/tex]
[tex]Z = -0.68[/tex]
[tex]Z = -0.68[/tex] has a pvalue of 0.2483
0.7517 - 0.2483 = 0.5034
0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months
