Suppose that blood chloride concentration(mmol/l) has a normal distribution with mean 110 and standard deviation 4. a. find the probability that chloride concentration equals 111= is less than 111 = is at most 111 = . round to 4 decimal places. b. what is the probability that chloride concentration differs from the mean by more than 2.5 standard deviations? round to 4 decimal places. does this probability depend on the values of μ and σ?

a) P(X = 111) = 0; P(X < 111) = 0.5987; P(X ≤ 111) = 0.5987; b) 0.0124; No

Step-by-step explanation:

For part a, we use z scores.  The formula for the z score of an individual value is

Our mean, μ, is 110 and our standard deviation, σ, is 4.

The probability that x is equal to a given value is 0, regardless of the value, the mean or the standard deviation.  This is because the probability in a normal distribution is the area under the curve; this means there must be a range of numbers.

To find P(X < 111),

z = (111-110)/4 = 1/4 = 0.25

Using a z table, we see that the area under the curve to the right of this is 0.5987.

For P(X ≤ 111), we use the same probability as P(X < 111).  There is no distinction between "less than" and "less than or equal to."

For part b,

2.5 standard deviations above the mean is z = 2.5.  2.5 standard deviations below the mean is z = -2.5.

Using the z table, the area under the curve to the right of z = 2.5 is 0.9938.  The area under the curve to the right of z = -2.5 is 0.0062.  This makes the area between them

0.9938-0.0062 = 0.9876.  This means anything farther from the mean than this is

1-0.9876 = 0.0124.  This does not have anything to do with the value of the mean or the standard deviation; this is because we already have the z score.

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