Distribution

Question

Sheila's doctor is concerned that she may suffer from gestational diabetes (high blood glucose levels during pregnancy). There is variation both in the actual glucose level and in the blood test that measures the level. A patient is classified as having gestational diabetes if the glucose level is above 140 milligrams per deciliter (mg/dl) one hour after a sugary drink is ingested. Sheila's measured glucose level one hour after ingesting the sugary drink varies according to the Normal distribution with μ = 122 mg/dl and σ = 10 mg/dl. What is the level L such that there is probability only 0.05 that the mean glucose level of 2 test results falls above L for Sheila's glucose level distribution? (Round your answer to one decimal place.)

 

(answer)mg/dl

Answer 1

For this we are using the mean, SD and the z table. The z table (look at the positive side) will show you the z value for an event that has less than a 5% probability (0.05) of being higher than that number. So first, recognize that the normal distribution with z scores of 1, 2,3 , 0 in the middle, and -1, -2, -3 represents a total probability of 100% under the curve. Look at the z score of 0. 50% of the area is to the left and 50% is to the right. For these data, 122 mg/dl is at the middle. Since SD is 10, you can plot some of the other values. So a blood glucose of 122 + 10 = 132 has a z score of +1, and a blood glucose of 122 + 20 = 142 has a z score of 2. If you look on the table, find the z score that corresponds to an area of 0.95. (It will be to many decimal places.) It should be 1.64. 1.6 will be on the left, then go over to the right  and see the number under 1.64. Now you are going to back calculate to the blood level using the formula z = (x - µ)/SD. z = 1.64, x is unknown, µ is 122, and SD is 10. Solve for x. This is the value of blood glucose level L.