The Intelligence Quotient (IQ) test scores for adults are normally distributed with a mean of 100 and a standard deviation of 15. What is the probability we could select a sample of 50 adults and find the mean of this sample is between 95 and 105?
0.0091
To find the probability of selecting a sample of 50 adults with a mean between 95 and 105, we will use the Central Limit Theorem.
The Central Limit Theorem states that for a large enough sample size, the distribution of sample means approaches a normal distribution, regardless of the shape of the population distribution. In this case, we have a large enough sample size (n=50) for the Central Limit Theorem to apply.
First, we need to calculate the standard error of the mean (SEM), which is the standard deviation of the sample mean. The formula for SEM is:
SEM = σ / √n
where σ is the population standard deviation and n is the sample size.
In this case, the population standard deviation (σ) is 15, and the sample size (n) is 50. Plugging these values into the formula, we get:
SEM = 15 / √50
≈ 2.12
Next, we need to convert the given range of means (95 to 105) into z-scores.
The formula for z-score is:
z = (x - μ) / σ
where x is the value of interest, μ is the population mean, and σ is the population standard deviation.
For the lower limit (95), the z-score is:
z1 = (95 - 100) / 2.12
≈ -2.36
For the upper limit (105), the z-score is:
z2 = (105 - 100) / 2.12
≈ 2.36
Now we need to find the corresponding probabilities for these z-scores using the standard normal distribution (z-distribution) table or calculator.
From the z-distribution table, we find that the area to the left of z1 (-2.36) is approximately 0.0099, and the area to the left of z2 (2.36) is also 0.9901.
To find the probability of the mean falling within this range, we subtract the lower probability from the upper probability:
Probability = 0.9901 - 0.0099
= 0.9802
Therefore, the probability of selecting a sample of 50 adults and finding the mean of this sample to be between 95 and 105 is approximately 0.9802 or 98.02%.
To find the probability of selecting a sample with a mean between 95 and 105, we need to calculate the z-scores for these values and then use the z-table to find the corresponding probabilities.
Step 1: Calculate the standard error of the sample mean.
The standard error of the sample mean, denoted as σ/√n, can be calculated by dividing the standard deviation (σ) by the square root of the sample size (n). In this case, σ = 15 and n = 50.
Standard Error = 15 / √50
Step 2: Calculate the z-scores for the mean of 95 and 105.
To calculate the z-score for a given value, we subtract the population mean from the value and then divide the result by the standard error.
For a mean of 95:
z-score = (95 - 100) / Standard Error
For a mean of 105:
z-score = (105 - 100) / Standard Error
Step 3: Use the z-table to find the corresponding probabilities.
The z-table provides the area under the standard normal distribution curve up to a certain z-score. We can use it to find the probabilities associated with the calculated z-scores.
P(95 ≤ X ≤ 105) = P(z-score1 ≤ Z ≤ z-score2)
Using the z-table, find the values corresponding to the z-scores calculated in Step 2 and subtract the area under the curve corresponding to the lower z-score from the area corresponding to the higher z-score.
Step 4: Calculate the probability.
Subtract the area under the curve corresponding to the lower z-score from the area corresponding to the higher z-score to get the probability.
Note: Depending on the available z-table or software, you may need to convert the z-scores to cumulative probabilities before finding the area between the two z-scores.