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

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

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(x-\mu)/((\sigma)/(√(n)))`

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

P(x < 2)

`P( x < 2) = P( z < \displaystyle(2-2.5)/((0.1)/(√(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.