Aperson'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 = 85 and standard deviation σ = 24. 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. (a) x is more than 60

(b) x is less than 110

(c) x is between 60 and 110

(d) x is greater than 140 (borderline diabetes starts at 140)

(a) 0.8512 (b) 0.8512 (c) 0.7024 (d) 0.0110

Step-by-step explanation:

The blood glucose follows a normal distribution N(μ=85;σ=24).

For every value of X, we can calculate the z-score (equivalent for a N(0;1)) and compute the probability.

(a) P(x>60)

z = (x-μ)/σ = (60-85)/24 = -1.0417

P(x>60) = P(z>-1.0417) = 0.8512

(b) P(x<110)

z = (x-μ)/σ = (110-85)/24 = 1.0417

P(x<110) = P(z<1.0417) = 0.8512

(c) P(60<x<110) = P(x<110)-P(x<60)

P(60<x<110) = P(z<1.0417) - P(z<-1.0417)

P(60<x<110) = 0.8512 - (1-0.8512) = 0.8512 - 0.1488 = 0.7024

(d) P(x>140)

z = (x-μ)/σ = (140-85)/24 = 2.2917

P(x>140) = P(z>2.2917) = 0.0110

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

1696 hundred thousand

step-by-step explanation: