The mean weight of loads of rock is 51.0 tons with a standard deviation of 10.0 tons. If 36 loads are chosen at random for a weight check, find the probability that the mean weight of those loads is less than 50.3 tons. Assume that the variable is normally distributed.

0.3372

Step-by-step explanation:

Mean weight (μ) = 51.0 tons

Standard deviation (σ) = 10.0 tons

n = 36

Pr (x<50.3) = ?

Using normal distribution,

Z = (X - μ) / σ/√n

When X = 50.3

Z = (50.3 - 51.0) / 10/√36

Z = - 0.7 / 10/6

Z = - 0.42

From the normal distribution table, 0.42 = 0.1628

Φ (z) = 0.1628

Recall that when Z is negative, Pr (x
Pr (x < 50.3) = 0.5 - 0.1628

= 0.3372