The average age of statistics students nationwide is 22. The standard deviation is 2.5 years. Assume the age is a normally distributed variable.
Find the probability that one student selected at random is older than 23.
Find the probability that the mean age of a group of 16 students selected at random is bigger than 23
Z = (score - mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion above that Z score.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
Since only one SD is provided, you can use just that to determine SEdiff.
Use same table.
To find the probability that one student selected at random is older than 23, we will use the z-score formula and the standard normal distribution.
Step 1: Calculate the z-score.
The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
x = the value we want to find the probability for (in this case, 23)
μ = the mean of the distribution (in this case, 22)
σ = the standard deviation of the distribution (in this case, 2.5)
Plugging the values into the formula:
z = (23 - 22) / 2.5
z = 1 / 2.5
z = 0.4
Step 2: Find the probability using the z-score.
To find the probability corresponding to the z-score of 0.4, we can use a standard normal distribution table (also known as a z-table) or use a calculator with the normal distribution function. The z-table provides the area under the curve up to a specific z-score.
Consulting the z-table, we find that the area to the left of 0.4 is approximately 0.6554. Since the area under the entire curve is equal to 1, the area to the right of 0.4 (representing being older than 23) is equal to 1 - 0.6554 = 0.3446.
Therefore, the probability that one student selected at random is older than 23 is approximately 0.3446, or 34.46%.
To find the probability that the mean age of a group of 16 students selected at random is bigger than 23, we will use the sampling distribution of the sample mean, assuming the population follows a normal distribution.
The mean of the sampling distribution of the sample mean (also known as the mean of the sampling distribution) is equal to the population mean, which is 22 in this case.
The standard deviation of the sampling distribution of the sample mean (also known as the standard error of the mean) can be calculated using the formula:
σ(¯x) = σ / √n
Where:
σ = population standard deviation (in this case, 2.5)
n = sample size (in this case, 16)
Plugging the values into the formula:
σ(¯x) = 2.5 / √16
σ(¯x) = 2.5 / 4
σ(¯x) = 0.625
Next, we need to calculate the z-score for the given mean age of 23 using the formula:
z = (x - μ(¯x)) / σ(¯x)
Where:
x = the value we want to find the probability for (in this case, 23)
μ(¯x) = the mean of the sampling distribution of the sample mean (which is the same as the population mean, 22)
σ(¯x) = the standard error of the mean (0.625)
Plugging the values into the formula:
z = (23 - 22) / 0.625
z = 1.6
Finally, we can use the z-score of 1.6 to find the probability by referring to the z-table. The area to the left of 1.6 is approximately 0.9452. Since we want to find the probability of the mean age being bigger than 23, we need to subtract this value from 1:
1 - 0.9452 = 0.0548
Therefore, the probability that the mean age of a group of 16 students selected at random is bigger than 23 is approximately 0.0548, or 5.48%.