Answer:
(a) Yes because the z-score is 2.5
Step-by-step explanation:
To solve this problem, 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.
Z-scores higher than 2 or lower than -2 are considered unusual
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 [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 40, \sigma = 4, n = 25, s = \frac{4}{\sqrt{25}} = 0.8[/tex]
Would a sample mean of 42 be unusual?
We have to find Z when X = 42. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{42 - 40}{0.8}[/tex]
[tex]Z = 2.5[/tex]
A z-score of 2.5 means that a sample mean of 42 would be unusual.
So the correct answer is:
(a) Yes because the z-score is 2.5
Answer:
(a) Yes because the z-score is 2.5
Step-by-step explanation:
z = (42 - 40)/(4/sqrt(25))
z = 2/0.8 = 2.5
|z| > 2 so unusual
