5. Among adults in the United States, the albumin levels (albumin is a type of p

Question

5. Among adults in the United States, the albumin levels (albumin is a type of protein) in cerebrospinal fluid is approximately normal with mean 29.5mg/100ml and standard deviation of 9.25mg/100ml.

a. What is the probability that the albumin level of a randomly selected adult in the population is between 29.5mg/100 ml and 46mg/100ml?

b. In this population, what is the albumin level so that only 3% of the population has a higher level?

c. Select 30 adults at random from this population. What is the probability that the average albumin level of the sample is between 30mg/100ml and 32mg/100ml?

d. Suppose the average albumin level of the 30 randomly selected adults from the population is above 30mg/100ml, what is the probability that the average albumin level is less than 32mg/100ml?

Explanation / Answer

a)

Mean = 29.5

Standard deviation = 9.25

Z1 = (X1 - mean)/(Standard deviation) = (29.5-29.5)/9.25 = 0

Z2 = (X2-mean)/(Standard deviation) = (46-29.5)/(9.25) = 1.781

P(29.5 < X < 46) = P(Z < 1.781) - P(Z<0) = 0.963 - 0.500 = 0.463

b)

Z-value for probability (0.03) = 1.881

Z = (X-Mean)/(standard deviation)

1.881 = (X-29.5)/(9.25)

X = 29.5 + 9.25 * 1.881 = 46.89

c) new SD = SD/sqrt(n) = 9.25/sqrt(30) = 1.6888

Z1 = (30 - 29.5)/(1.6888) = 0.2960

Z2 = (32 - 29.5)/(1.6888) = 1.480

P(30<X<32) = P(Z< 1.480) - P(Z<0.2960) = 0.931 - 0.616 = 0.315

d)

Probability = (Z2 int Z1)/Z1 = (0.931-0.616)/(0.616) = 0.5113

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