Spring 20186 Quiz-2 deviation of 40 mg/dl. s normally distributed with a mean sc

Question

Spring 20186 Quiz-2 deviation of 40 mg/dl. s normally distributed with a mean score of 180 mg/dl and a standard ance person is picked at random and tested. Find the probability that the person has (1) Cholesterol value less than (2) Cholesterol value more than (3) Cholesterol value between 100 and 260 mg/d 100 mg/dl 260 mg/dl (4) Wh (5) Wh at is the cholesterol value that will put a person in the bottom 5.0 % at is the cholesterol value that will put a person in the top 2.5 % 6 100-180-2 - 5 30-60 4o

Explanation / Answer

Here, We will calculate Z for each of the situations and take the probability values from standard normal distribution table.

(1). Z = (100-180)/40 = -80/40 = -2

P(X < Z ) = 0.02275

(2) Z = (260-180) /40 = 80/40 = 2

P(X>Z) = 0.02275

(3) Z1 = -2 and Z2 = 2

P (Z1 < X < Z2) = 1-( 0.02275 + 0.02275 ) = 1-0.0455 = 0.9545

(4) To be in the bottom 5% , X should be less than a cutoff Z

cutoff Z is equal to -1.65

cutoff X will be equal to = Z * std. dev + mean = -1.65 * 40 + 180 = -66 + 180 =114

thus cholestrol less than 114 will put anyone in the bottom 5%

(5) to be in top 2.5%, cutoff Z is equal to 1.96

cutoff X is equal to = 1.96 * 40 + 180 = 78 + 180 = 258

Thus cholestrol greater than 258 will put anyone in top 2.5%

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