Question

Miguel buys a large bottle and a small bottle of juice. The amount of juice that the manufacturer puts in the large bottle is a random variable with a mean of 1016 ml and a standard deviation of 8 ml. The amount of juice that the manufacturer puts in the small bottle is a random variable with a mean of 510 ml and a standard deviation of 5ml. If the total amount of juice in the two bottles can be described by a normal model, what’s the probability that the total amount of juice in the two bottles is more than 1540.2 ml?

Answer 1

Answer:

Pr(X>1540.2) = 0.0655

Step-by-step explanation:

Expected value of large bottle,

E(Large) = 1016

Expected value of small bottle,

E(small) = 510

Expected value of total

E(total) = 1016 + 510 = 1526

So the new mean is 1526

Find standard deviation of new amount by variance

Variance of large bottle,

v(large) = 8^2 = 64

Variance of small bottle,

v(small) = 5^2 = 25

Variance of total

v(total) = 64+25 = 89

So the new standard deviation

sd(new) = sqrt(89) = 9.434

Find probability using the new mean and s.d.

Pr(X>1540.2)

Z score, z = (x-mean)/sd

= (1540.2 - 1526)/9.434

= 1.505

value in z score

P(z<1.51) = 0.9345

For probability of x > 1540.2

P(z > 1.51) = 1 - 0.9345 = 0.0655