A worldwide organization of academics claims that the mean IQ score of its members is 118, with a standard deviation of 17. A randomly selected group of 40 members of this organization is tested, and the results reveal that the mean IQ score in this sample is 115.8. If the organization's claim is correct, what is the probability of having a sample mean of 115.8 or less for a random sample of this size

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 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 [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.

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 [tex]\mu = 118, \sigma = 17[/tex]

A randomly selected group of 40 members

This means that [tex]n = 40, s = \frac{17}{\sqrt{40}} = 2.6879[/tex]

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.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{115.8 - 118}{2.6879}[/tex]

[tex]Z = -0.82[/tex]

[tex]Z = -0.82[/tex] 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

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