Question

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

Answer 1

`P(\bar X <117.2)`

And we can use the z score:

`z= (\bar X -\mu)/((\sigma)/(√(n)))`

And if we find the z score for 117.2 we got:

`z = (117.2-112)/((15)/(√(35))) = 2.0509`

And we can find the probability using the normal standard table or excel and we got:

`P(\bar X <117.2)= P(z<2.0509) = 0.980`

Explanation:

Assuming this problem: "A worldwide organization of academics claims that the mean IQ score of its members is 112, with a standard deviation of 15. A randomly selected group of 35 members of this organization is tested, and the results reveal that the mean IQ score in this sample is 117.2. If the organization's claim is correct, what is the probability of having a sample mean of 117.2 or less for a random sample of this size?

Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places."

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Let X the random variable that represent the IQ scores of a population, and for this case we know the following properties:

`E(X)= 112, \sigma = 15`

From the central limit theorem we know that the distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

And for this case we want to find this probability:

`P(\bar X <117.2)`

And we can use the z score:

`z= (\bar X -\mu)/((\sigma)/(√(n)))`

And if we find the z score for 117.2 we got:

`z = (117.2-112)/((15)/(√(35))) = 2.0509`

And we can find the probability using the normal standard table or excel and we got:

`P(\bar X <117.2)= P(z<2.0509) = 0.980`