3. Research indicates that the concentration of chlorides in blood (measured in

Question

3. Research indicates that the concentration of chlorides in blood (measured in mmol/L) has a normal distribution with mean 104 and a standard deviation of 5 (a) What is the probability that chloride concentration 112 ls less than 112? ls at moot 112? b) What is the probability that a blood chloride concentration is exceeds 100? (c) What is the probability that a blood chloride concentration is in between 90 and 120? (d) The highest 1% of blood chloride concentrations must be above what value? (e) The lowest 5% of blood chloride concentrations must be below what value?

Explanation / Answer

Quesrtion 3

Here mean choloorine level = ? = 104 mmOl/l

Standard deviation of chlorine level = ? = 5 mMl.litre

Here if x is the chlorine leve of blood of a ranodm person.

(i) Pr(x = 112) = NORMDIST(x = 112 ; 104; 5 ; False) = 0.0222

Pr(x < 112) = NORM(x < 112 ; 104 ; 5)

Z = (112 - 104)/5 = 8/5 = 1.6

Pr(x < 112) = NORM(x < 112 ; 104 ; 5) = Pr(Z< 1.6)

seeing in the table

Pr(x < 112) = NORM(x < 112 ; 104 ; 5) = Pr(Z< 1.6) = 0.9452

Pr(x > 112) = 1 - Pr(Z < 1.6) = 1 - 0.9452 = 0.0548

(ii) Pr(X > 100) = 1 - Pr(x < 100 ; 104; 5)

Z = (100 - 104)/5 = -0.8

Pr(x > 100) = 1 - Pr(x < 100 ; 104; 5) = 1 - Pr(Z < -0.8) = 1 - 0.2119 = 0.7881

(iii) Pr(90 < x < 120) = Pr(x < 120) - Pr(x < 90)

Z2 = (120 - 104)/.5 = 3.2 ; Z1 = (90 - 104)/5 = -2.8

Pr(90 < x < 120) = Pr(x < 120) - Pr(x < 90) = Pr(Z < 3.2) - Pr(Z < -2.8) = 0.9993 - 0.0026 = 0.9968

(iv)

Here we are talking about 99% percentile. Let say it is c

Pr(x > c) = NORM (x > c ; 104 ; 5) = 0.01

NORM (x < c ; 104 ; 5) = 1 - 0.01 = 0.99

the requisite Z - value from the table

Z = 2.326

(c - 104)/5 = 2.326

c = 104 + 5 * 2.326 = 115.63

(e)

Here let say that value is k

so,

Pr(x < k ; 104 ; 5) = 0.05

p - value = 1.645

(k - 104)/5 = 1.645

k = 104 + 5 * 1.645 = 112.225

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