18. The Intelligence Quotient (IQ) test scores are normally distributed with a mean of 100 and a standard deviation of 15. Given a class with 25 students, what is the probability that the class' average IQ score is between 85 and 115?
A) 0.0026
B) 0.4987
C) 3.00
D) 0.9974
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores.
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To solve this question, we need to use the Central Limit Theorem, which states that the distribution of sample means will be approximately normally distributed, regardless of the shape of the population, as long as the sample size is large enough (usually greater than 30).
The mean of the sample means will be equal to the population mean, which is 100 in this case. The standard deviation of the sample means, also known as the standard error, can be calculated by dividing the population standard deviation by the square root of the sample size.
In this case, the sample size is 25 and the population standard deviation is 15, so the standard error is:
Standard Error = 15 / √25
= 15 / 5
= 3
To find the probability that the class' average IQ score is between 85 and 115, we need to find the z-scores for these values. The z-score can be calculated using the formula:
z-score = (x - μ) / σ
where x is the given value, μ is the population mean, and σ is the population standard deviation.
For 85:
z-score = (85 - 100) / 3
= -15 / 3
= -5
For 115:
z-score = (115 - 100) / 3
= 15 / 3
= 5
Now, we need to find the probability that the z-score is between -5 and 5. We can use a standard normal distribution table to find this probability.
Using the table, we find that the probability of a z-score between -5 and 5 is approximately 0.9974.
Therefore, the correct answer is D) 0.9974.
To find the probability that the class' average IQ score is between 85 and 115, we need to use the Central Limit Theorem.
The Central Limit Theorem states that the sample mean of a large enough sample size from any population with a finite mean and standard deviation will be approximately normally distributed.
In this case, we are given that the IQ test scores are normally distributed with a mean of 100 and a standard deviation of 15. Since we have a sample size of 25 students, which is relatively large, we can use the Central Limit Theorem.
To calculate the probability, we need to convert the IQ scores to z-scores. The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
z is the z-score,
x is the value of interest (85 and 115),
μ is the mean of the distribution (100),
σ is the standard deviation of the distribution (15).
First, let's calculate the z-score for 85:
z1 = (85 - 100) / 15 = -1
Next, let's calculate the z-score for 115:
z2 = (115 - 100) / 15 = 1
Now, we need to find the probability corresponding to these z-scores using a standard normal distribution table or a calculator.
On a standard normal distribution table, the probability corresponding to z = -1.00 is approximately 0.1587, and the probability corresponding to z = 1.00 is also approximately 0.1587.
To find the probability of the class' average IQ score being between 85 and 115, we need to subtract the probability of z < -1.00 from the probability of z < 1.00:
P(-1.00 < z < 1.00) = P(z < 1.00) - P(z < -1.00)
= 0.1587 - 0.1587
= 0.0000
The probability is 0.0000.
Therefore, the correct answer is A) 0.0026.