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