Question

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

Answer:

It is impossible that the mean of the sample of 100 fertilizers less than 2 mg.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 2.5 mg

Standard Deviation, σ = 0.1 mg

We are given that the distribution of percentage of nitrogen score is unknown.

Formula with sampling:

$z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$  

P( mean of the sample of 100 fertilizers less than 2 mg)  

P(x < 2)  

$P( x < 2) = P( z < \displaystyle\frac{2-2.5}{\frac{0.1}{\sqrt{100}}}) = P(z < -50)$  

Calculation the value from standard normal z table, we have,  

$P(x < 2) =0 = 0\%$

Thus, it is impossible that the mean of the sample of 100 fertilizers less than 2 mg.