Answer the following questions.

Question

A person's blood glucose level and diabetes are closely related. Let x be a random variable measured in milligrams of glucose per deciliter (1/10 of a liter) of blood. Suppose that after a 12-hour fast, the random variable x will have a distribution that is approximately normal with mean 𝜇 = 86 and standard deviation 𝜎 = 23. Note: After 50 years of age, both the mean and standard deviation tend to increase. For an adult (under 50) after a 12-hour fast, find the following probabilities. (Round your answers to four decimal places.)

(a)

x is more than 60

 

(b)

x is less than 110

 

(c)

x is between 60 and 110

 

(d)

x is greater than 125 (borderline diabetes starts at 125)

 

Answer 1

a. P(X>60) = P(Z>(60-86)/23) = P(Z>-1.1304) = 1- P(X<-1.1304) = 1-0.1292 = 0.8708

b. P(X<110) = P(Z<(110-86)/23) = P(Z<1.0435) = 0.8516

c. P(60 < X < 110) = P(X<110) - P(X< 60) = 0.8516 - 0.1292 = 0.7224

d. P(X>125) = P(Z>(125-86)/23) = P(Z>1.6957) = 1- P(X<1.6597) = 1-0.9515 = 0.0485

Let me know if you have any questions. These problems can also be solved using the normalcdf function on the TI84

Dian

Answer 2

Hi Tamera,

(a) We're back to our classic equation: z= (x-mu)/sigma. Here,

x=60

mu=86

sigma=23

z= (60-86)/23

z= -1.13

So look up a z-score table and search for -1.1 on the left, 0.03 on the top row:

P(Z< -1.13) = 0.1292

However, the question asked for glucose level more than 60. This is where you apply the Complement Rule or, as I call it, the "one minus trick." Essentially,

P(Z>a)= 1- P(Z<a), where a represents any real number,

So, for our problem:

P(Z > -1.13) = 1-0.1292 = 0.8708

(b) This is the same equation, z= (x-mu)/sigma

x=110

mu=86

sigma=23

z= (110-86)/23

z= 1.04

P(Z<1.04) = 0.8508

Note that the question asked for less than, which is what the z-table provides, so we do not need to use the "one minus trick."

(c) Again, the same equation, but we have two values for x, which means we need two z-scores. I will refer to these as z₁ and z₂.

x₁ = 60

x₂ = 110

mu = 86

sigma = 23

z₁ = (60-86)/23

z₂ = (110-86)/23

z₁ = -1.13

z₂ = 1.04

Now, recall that the z-table gives only less than probabilities. Still:

P(z< -1.13) = 0.1292

P(z< 1.04) = 0.8508

Now, any time you are asked to find a range of normal probabilities of the form a < x < b, a and b real numbers, you subtract the smaller from the larger. Here:

P = 0.8508 - 0.1292 = 0.7216

(d) This is similar to (a). We want P(X>125), so we use our classic equation again:

z= (x - mu)/sigma

x=125

mu=86

sigma=23

z= (125-86)/23

z=1.70

P(Z<1.70)=0.9554

Again, remember that the question asked for greater, which means we need the "One Minus Trick:"

P(Z>1.70) = 1- P(Z<1.70)

P= 1 - 0.9554 = 0.0446

I hope this helps and, again, I have extensive experience with the normal distribution and z-scores, so if you want to set up a time to practice more examples with me, please do.