Respuesta :
Answer:
A sample size of 18.
Step-by-step explanation:
To solve this question, we have 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 random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], 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]
Combining them:
The formula for the z-score is:
[tex]Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In this problem, we have that:
[tex]\mu = 500, \sigma = 100[/tex]
For what sample size would you expect a sample mean of 489 to be at the 33rd percentile?
This is n as such Z has a pvalue of 0.33 when X = 489. So when X = 489, Z = -0.47.
So
[tex]Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]-0.47 = \frac{489 - 500}{\frac{100}{\sqrt{n}}}[/tex]
[tex]-0.47*100 = -11\sqrt{n}[/tex]
[tex]11\sqrt{n} = 47[/tex]
[tex]\sqrt{n} = \frac{47}{11}[/tex]
[tex]\sqrt{n} = 4.27[/tex]
[tex]\sqrt{n}^{2} = (4.27)^{2}[/tex]
[tex]n = 18[/tex]
A sample size of 18.
