Please help, I do not understand how to do this:
Let x be a random variable that represents white blood cell count per cubic milliliter of whole blood. Assume that x has a distribution that is approximately normal, with mean μ = 7400 and estimated standard deviation σ = 2900. A test result of x < 3500 is an indication of leukopenia. This indicates bone marrow depression that may be the result of a viral infection.
(a) What is the probability that, on a single test, x is less than 3500? (Round your answer to four decimal places.)
(b)What is the probability of x < 3500? (Round your answer to four decimal places.)
(c) Repeat part (b) for n = 3 tests taken a week apart. (Round your answer to four decimal places.)
(A) From the problem statement, N(7400, 2900).
P(X < 3500) = .0893
(B) ... .0893?
(C)
This is a test of means. For this we make a new distribution, N(xbar, s), where xbar = mu, s = sigma / sqrt(n).
N(7400, 1674.32)
P(X < 3500) = .0099
Note: For part c, the probability statement should read P(xbar < 3500). The bar denotes sampling means.
Thank you so much, I appreciate it!
To find the probability that a random variable is less than a certain value, we can use the standard normal distribution.
(a) To find the probability that x is less than 3500 on a single test, we need to standardize the value using the z-score formula: z = (x - μ) / σ. In this case, the z-score is (3500 - 7400) / 2900 ≈ -1.206.
Next, we can use the z-table or a calculator to find the corresponding probability. From the z-table, the probability corresponding to -1.206 is approximately 0.1131.
Therefore, the probability that x is less than 3500 on a single test is approximately 0.1131.
(b) The probability of x < 3500 is the same as the probability that x is less than or equal to 3500. Using the z-table or calculator, we find that the probability corresponding to -1.206 is approximately 0.1131.
Therefore, the probability of x < 3500 is approximately 0.1131.
(c) To find the probability of x < 3500 for n = 3 tests taken a week apart, we need to consider the independent nature of each test.
Since each test is taken a week apart, we can assume that they are independent random variables with the same mean and standard deviation. The sum of independent normal random variables is also a normal random variable.
The mean of the sum of n independent random variables is n times the mean of the individual random variable. In this case, the mean of the sum is 3(7400) = 22200.
The standard deviation of the sum of n independent random variables is the square root of the sum of the variances of the individual random variables. In this case, the standard deviation of the sum is sqrt(3(2900)^2) ≈ 5011.
Now we can standardize the value of 3500 using the z-score formula: z = (3500 - 22200) / 5011 ≈ -3.433.
Using the z-table or calculator, we find that the probability corresponding to -3.433 is approximately 0.0003.
Therefore, the probability of x < 3500 for n = 3 tests taken a week apart is approximately 0.0003.