5. A person’s blood glucose level and diabetes are closely. X is a random variable that measures the milligrams of glucose per deciliter of blood. After a 12 hour fast, x has an approximate normal distribution with a mean of 85 and a standard deviation of 25. What is the probability in a. through d) that :
a. X is more than 60?
b. X is less than 110?
c. X is between 60 and 110
d. X is greater than 140 (borderline diabetes starts at 140)?
e. What would the minimum X value for blood glucose be if someone ranked in the highest 25% for blood glucose level?
a - d. Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities related to the Z scores.
e. Reverse process, starting with .25 and its Z score
To find the probabilities mentioned in the question, we need to use the properties of the normal distribution and the given parameters of the random variable X.
The normal distribution can be standardized using the formula: Z = (X - μ) / σ, where Z is the standardized value, X is the observed value, μ is the mean, and σ is the standard deviation.
a) To find the probability that X is more than 60, we need to calculate P(X > 60).
First, we standardize the value 60 using the formula: Z = (60 - 85) / 25 = -1.
Next, we look up the corresponding area to the left of Z = -1 in the standard normal distribution table, which is approximately 0.1587.
However, we want the probability of X being greater than 60, so we subtract the area from 1:
P(X > 60) = 1 - 0.1587 = 0.8413
Therefore, the probability that X is more than 60 is 0.8413 or 84.13%.
b) To find the probability that X is less than 110, we need to calculate P(X < 110).
First, we standardize the value 110 using the formula: Z = (110 - 85) / 25 = 1.
Next, we look up the corresponding area to the left of Z = 1 in the standard normal distribution table, which is approximately 0.8413.
P(X < 110) = 0.8413
Therefore, the probability that X is less than 110 is 0.8413 or 84.13%.
c) To find the probability that X is between 60 and 110, we need to calculate P(60 < X < 110).
We have already found the individual probabilities in parts a) and b).
P(X < 60) = 0.1587 (from part a)
P(X < 110) = 0.8413 (from part b)
To find the probability in the given range, we subtract these probabilities:
P(60 < X < 110) = P(X < 110) - P(X < 60) = 0.8413 - 0.1587 = 0.6826
Therefore, the probability that X is between 60 and 110 is 0.6826 or 68.26%.
d) To find the probability that X is greater than 140, we need to calculate P(X > 140).
First, we standardize the value 140 using the formula: Z = (140 - 85) / 25 = 2.2.
Next, we look up the corresponding area to the left of Z = 2.2 in the standard normal distribution table. However, the table only provides values up to 2.0.
Since we know that the standard normal distribution is symmetric, we can approximate the area using the value for Z = 2.0, which is approximately 0.9772.
Therefore, P(X > 140) ≈ 1 - 0.9772 = 0.0228
The probability that X is greater than 140 is approximately 0.0228 or 2.28%.
e) To determine the minimum X value for blood glucose if someone ranks in the highest 25%, we need to find the Z-score for which the area to the left is 0.75 (1 - 0.25).
Looking up the area 0.75 in the standard normal distribution table, we find that the corresponding Z-score is approximately 0.674.
Using the formula for standardizing, we can solve for X:
Z = (X - μ) / σ
0.674 = (X - 85) / 25
Simplifying the equation, we find:
0.674 * 25 = X - 85
16.85 + 85 = X
X ≈ 101.85
Therefore, the minimum X value for blood glucose if someone ranks in the highest 25% would generally be approximately 101.85 milligrams per deciliter of blood.