Question

The actual weights of bags of pet food are normally distributed with a mean weight of a bag of 50.0 Ib., and a standard deviation of 0.2 lb. e) If there is a 35% chance to choose a bag with weight greater or equal to X what is X2

Answer 1

c) To find the desired probability, we will convert the random variable X, which indicates the weight of the bags, into the standard normal variable z. The formula used is

$z = \frac{X - \mu}{\sigma}$

$P(X > 50.3) = P(\frac{X - \mu}{\sigma} > \frac{50.3 - \mu}{\sigma}) \\ = P(z > \frac{50.3 - 50}{0.2}) = P(z > 1.5) \\ = 1 - P(z \leq 1.5) = 1 - 0.93319 = 0.06681$

Expected number of bags over 50.3 = 250*0.06681 = 16.7025

d)

$P(X > 50.15) = P(\frac{X - \mu}{\sigma} > \frac{50.15 - \mu}{\sigma}) \\ = P(z > \frac{50.15 - 50}{0.2}) = P(z > 0.75) \\ = 1 - P(z \leq 0.75) = 1 - 0.77337 = 0.22663$

The chances of a bag weighing more than 50.15 is 0.22663.

e)

$P(X \geq X_1) = 0.35 \\ P(z \geq z_1) = 0.35 \\ 1 - P(z \leq z_1) = 0.35 \\ P(z \leq z_1) = 0.65 \\ z_1 = 0.385 \\ X_1 = \mu + z_1 \sigma = 50 + 0.385*0.2 = 50.077$