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?

Answer:a) P(X = 111) = 0; P(X < 111) = 0.5987; P(X ≤ 111) = 0.5987; b) 0.0124; NoStep-by-step explanation:For part a, we use z scores.  The formula for the z score of an individual value is[tex]z=\frac{X-\mu}{\sigma}[/tex]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 = 0.25Using 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 them0.9938-0.0062 = 0.9876.  This means anything farther from the mean than this is1-0.9876 = 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.