Systolic blood pressure for women between the ages of 18 and 24 is normally distributed with a mean of 114.8 (millimeters of mercury) and a standard deviation of 13.1.
A. What is the likelihood that an individual woman has a blood pressure abouve 125?
B. Suppose a random sample of n=300 women is selected and the mean blood pressure for tyhe sample is computed. What is the likelihood that the mean blood pressure for the sample will be above 125?
C. Suppose a random sample of n=300 women is selected and the mean blood pressure for the sample is computed. What is the likelihood that the mean blood pressure for the sample will be bleow. 114?
A. Z = (score-mean)/SD (for distribution of scores)
B & C. Z = (mean1 - mean2)/standard error (SE) of difference between means (for distribution of means)
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
Since only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to these Z scores.
To find the likelihood or probability in these scenarios, we need to use the concept of the standard normal distribution. The standard normal distribution has a mean of 0 and a standard deviation of 1. We can convert our given information about the systolic blood pressure of women to the standard normal distribution using the formula:
Z = (X - μ) / σ
Where:
Z is the standard score or z-score,
X is the value we want to convert,
μ is the population mean, and
σ is the population standard deviation.
A. What is the likelihood that an individual woman has a blood pressure above 125?
To find the likelihood that an individual woman has a blood pressure above 125, we need to convert this value to a z-score using the given mean and standard deviation:
Z = (125 - 114.8) / 13.1
Z = 0.765
Using a standard normal distribution table or calculator, we can find the probability corresponding to this z-score. The table provides the probability from the left tail to the z-score. Since we need the probability above 125, we subtract the probability obtained from 1:
P(X > 125) = 1 - P(Z < 0.765)
By referring to a standard normal distribution table, we find that P(Z < 0.765) is approximately 0.7764. Therefore:
P(X > 125) = 1 - 0.7764
P(X > 125) ≈ 0.2236
So, the likelihood that an individual woman has a blood pressure above 125 is approximately 0.2236 or 22.36%.
B. What is the likelihood that the mean blood pressure for the sample will be above 125?
To find the likelihood that the mean blood pressure for a sample of 300 women will be above 125, we need to use the concept of the sampling distribution of the sample mean. The sampling distribution of the sample mean is approximately normally distributed, with the same mean as the population mean but with a standard deviation equal to the population standard deviation divided by the square root of the sample size.
First, we find the standard deviation of the sampling distribution:
σ_sample_mean = σ / √(n)
σ_sample_mean = 13.1 / √300
σ_sample_mean ≈ 0.757
Next, we convert the value 125 to a z-score using the known population mean and the calculated standard deviation of the sample mean:
Z = (125 - 114.8) / 0.757
Z ≈ 13.5
Since the sample size is relatively large (n = 300), we can use the standard normal distribution to find the probability of obtaining a sample mean above 125:
P(X_bar > 125) = P(Z > 13.5)
Referring to a standard normal distribution table or using a calculator, we find that P(Z > 13.5) is an extremely small value (close to 0). Therefore, the likelihood that the mean blood pressure for the sample will be above 125 is negligible.
C. What is the likelihood that the mean blood pressure for the sample will be below 114?
We can follow the same steps as in part B to find the likelihood that the mean blood pressure for the sample will be below 114. Converting 114 to a z-score using the standard deviation of the sample mean:
Z = (114 - 114.8) / 0.757
Z ≈ -1.06
P(X_bar < 114) = P(Z < -1.06)
By referencing a standard normal distribution table or calculator, we find that P(Z < -1.06) is approximately 0.1419. Therefore:
P(X_bar < 114) ≈ 0.1419
So, the likelihood that the mean blood pressure for the sample will be below 114 is approximately 0.1419 or 14.19%.