In a large population of college-educated adults, the mean IQ is 118 with a standard deviation of 20. You do not know the population distribution for the IQ scores - only the population mean and the population standard deviation. Suppose 200 adults from this population are randomly selected for a market research campaign.
What is the probability that the sample mean IQ is greater than 120?

A) 0.4602 B) 0.0793 C) 0.5398 D) 0.9207

B) 0.0793

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 and standard deviation , the zscore of a measure X is given by:

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 and standard deviation , a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error

In this problem, we have that:

What is the probability that the sample mean IQ is greater than 120?

This probability is 1 subtracted by the pvalue of Z when X = 120. So

has a pvalue of 0.9207

1 - 0.9207 = 0.0793

So the correct answer is:

B) 0.0793

it will cost dollar $540

the living room is 3 which means if we divide 325/3 then we get $108

therefore if 1part of the living room cost $108 and 3 costed $325

then total 5part of the room will cost 5*108 which gives $540

step-by-step explanation: it is d