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

Solution:

Given:

Mean
=μ = 2.5
mg
Standard Deviation =
0=0.1
mg
Sample size = n = 100

We have to find:

Plī<2) = ......

Since sample size n = 100 is large , we can use Central limit theorem which states that for large sample size n ,
sampling distribution of sample mean is approximately normal with mean of sample means:

Vi=

Mi = 2.5

and standard deviation of sample means is:

0i = n

0 =

0.1 = 40

0 = 0.01

Thus find z score:

1 - Hi 2=

2-2.5 0.01

-0.5 0.01

了= -50

Since z = -50 is too far below z = -3 or z = -50 is below 3 standard deviations from mean, thus probability of z below -50 is approximately 0

that is: P(z < -50) = 0.0000

thus

Plī <2) = P(Z < -50)

Plī<2) = 0.0000

Thus required answer is 0.0000