Answer:
0.2061 = 20.61% probability of having a sample mean of 115.8 or less for a random sample of this size
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 normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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 p-value, 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
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
A worldwide organization of academics claims that the mean IQ score of its members is 118, with a standard deviation of 17.
This means that ![\mu = 118, \sigma = 17](https://tex.z-dn.net/?f=%5Cmu%20%3D%20118%2C%20%5Csigma%20%3D%2017)
A randomly selected group of 40 members
This means that ![n = 40, s = \frac{17}{\sqrt{40}} = 2.6879](https://tex.z-dn.net/?f=n%20%3D%2040%2C%20s%20%3D%20%5Cfrac%7B17%7D%7B%5Csqrt%7B40%7D%7D%20%3D%202.6879)
What is the probability of having a sample mean of 115.8 or less for a random sample of this size?
This is the pvalue of Z when X = 115.8.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{115.8 - 118}{2.6879}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B115.8%20-%20118%7D%7B2.6879%7D)
![Z = -0.82](https://tex.z-dn.net/?f=Z%20%3D%20-0.82)
has a pvalue of 0.2061
0.2061 = 20.61% probability of having a sample mean of 115.8 or less for a random sample of this size