Question

QUESTION 18 A large cooperation has quality control over its fertilizers. The fertilizes are composed of nitrogen. The fertilizer requires 3 mg of nitrogen. The distribution of the percentage of nitrogen is unknown with a mean of 2.5 mg and a standard deviation of 0.1. A specialist randomly checked 100 fertilizer samples. What is the probability that the mean of the sample of 100 fertilizers less than 2 mg?

Answer 1

Solution :

Given that ,

mean ($\mu$) = 2.5

standard deviation ($\sigma$) = 0.1

n = 100

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 0.1 / $\sqrt$ 100 = 0.01

P($\bar x$ < 2) = P[($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$} < (2 - 2.5) /0.01]

= P(z < -50)

Probability = 0