In a large population of adults, the mean IQ is 112 with a standard deviation of 20. Suppose 200 adults are randomly selected for a market research campaign. The probability that the sample mean IQ is greater than 110 is...?
your z score is (110-112)/20 = -.1
if you have a normal distribution table, look up the probability for that score, you should get .4602
that means the the prob. of a score of less than 110 is .4602
so the prob of getting higher than 110 is 1-.4602
= .5398
here is a nice applet that lets you enter your original values, it graphs it and gives you the results
http://davidmlane.com/hyperstat/z_table.html
To find the probability that the sample mean IQ is greater than 110, we can use the Central Limit Theorem, which states that the distribution of the sample means approaches a normal distribution as the sample size increases.
First, let's calculate the standard deviation of the sample mean using the formula:
Standard deviation of the sample mean = Standard deviation of the population / Square root of the sample size
In this case, the standard deviation of the population is 20 and the sample size is 200. So, the standard deviation of the sample mean is:
Standard deviation of the sample mean = 20 / sqrt(200)
= 20 / 14.1421
≈ 1.4142
Next, we need to find the z-score for 110. The z-score measures the number of standard deviations a value is from the mean. It is calculated using the formula:
z-score = (x - mean) / standard deviation
where x is the value and mean is the mean of the distribution.
In this case, the mean is 112 and the standard deviation is 1.4142. So, the z-score for 110 is:
z-score = (110 - 112) / 1.4142
= -2 / 1.4142
≈ -1.414
Now, we need to find the probability that the z-score is greater than -1.414. We can look up this probability in a standard normal distribution table or use a calculator.
Using a standard normal distribution table or calculator, we find that the probability that the z-score is greater than -1.414 is approximately 0.9222.
Therefore, the probability that the sample mean IQ is greater than 110 is approximately 0.9222 or 92.22%.
To find the probability that the sample mean IQ is greater than 110, we can use the Central Limit Theorem.
The Central Limit Theorem states that regardless of the shape of the population distribution, the distribution of the sample means will approach a normal distribution as the sample size increases.
In this case, the population has a mean IQ of 112 and a standard deviation of 20. We are interested in finding the probability that the sample mean IQ is greater than 110.
First, let's calculate the standard error. The standard error is the standard deviation of the sample mean and is given by the formula:
Standard Error = Standard Deviation / √(Sample Size)
In this case, the standard deviation (σ) is 20 and the sample size (n) is 200. Plugging these values into the formula, we get:
Standard Error = 20 / √200
Calculating this, we find that the standard error is approximately 1.414.
Next, we need to standardize the sample mean using the standard error to convert it into a standard normal distribution. We can use the z-score formula:
z = (Sample Mean - Population Mean) / Standard Error
Plugging in the values, we get:
z = (110 - 112) / 1.414
Calculating this, we find that the z-score is approximately -1.414.
Now, we can use a z-table or a calculator to find the probability associated with the z-score. The probability represents the area under the normal curve to the right of the z-score.
Looking up the z-score in a standard normal distribution table, we find that the probability associated with this z-score is approximately 0.078.
Therefore, the probability that the sample mean IQ is greater than 110 is approximately 0.078, or 7.8%.