Respuesta :
Answer:
a) 0.0951
b) 0.8098
c) Between $24.75 and $27.25.
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 = 26, \sigma = 6, n = 62, s = \frac{6}{\sqrt{62}} = 0.762[/tex]
(a)
What is the likelihood the sample mean is at least $27.00?
This is 1 subtracted by the pvalue of Z when X = 27. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{27 - 26}{0.762}[/tex]
[tex]Z = 1.31[/tex]
[tex]Z = 1.31[/tex] has a pvalue of 0.9049
1 - 0.9049 = 0.0951
(b)
What is the likelihood the sample mean is greater than $25.00 but less than $27.00?
This is the pvalue of Z when X = 27 subtracted by the pvalue of Z when X = 25. So
X = 27
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{27 - 26}{0.762}[/tex]
[tex]Z = 1.31[/tex]
[tex]Z = 1.31[/tex] has a pvalue of 0.9049
X = 25
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{25 - 26}{0.762}[/tex]
[tex]Z = -1.31[/tex]
[tex]Z = -1.31[/tex] has a pvalue of 0.0951
0.9049 - 0.0951 = 0.8098
c)Within what limits will 90 percent of the sample means occur?
50 - 90/2 = 5
50 + 90/2 = 95
Between the 5th and the 95th percentile.
5th percentile
X when Z has a pvalue of 0.05. So X when Z = -1.645
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]-1.645 = \frac{X - 26}{0.762}[/tex]
[tex]X - 26 = -1.645*0.762[/tex]
[tex]X = 24.75[/tex]
95th percentile
X when Z has a pvalue of 0.95. So X when Z = 1.645
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]1.645 = \frac{X - 26}{0.762}[/tex]
[tex]X - 26 = 1.645*0.762[/tex]
[tex]X = 27.25[/tex]
Between $24.75 and $27.25.
