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Answer:
A. P(x<3.85 or x>4.15)= P(x<3.85)+P(x>4.15) = 0.1336
Step-by-step explanation:
Working with an ordinary Normal Distribution of probability and trying to find the probabilities asked in it could be difficult, because there´s no easy method to find probabilities in a generic Normal Distribution (with mean μ=4 and STD σ=0.1). The recommended approach to this question is to use a process called "Normalize", this process let us translate the problem of any Normal Distribution to a Standard Normal Distribution (μ=0 and σ=1) where there´s easier ways to find probabilities in there. The "Normalization" goes as follows:
Suppose you want to know P(x<a) of the Normal Distribution you are working with:
P(x<a)=P( (x-μ)/σ < (a-μ)/σ )=P(z<b) ( b=(a-μ)/σ )
Where μ is the mean and σ is the STD of your Normal Distribution. Notice P(z<b) now it´s a probability in a Standard Normal Distribution, now we can find it using the available method to do so. My favorite is a chart (It´s attached to this answer) that contains a lot of probabilities in a Standard Normal Distribution. Let´s solve this as an example
A. We want to find the probability of the cork being defective (P(x<3.85) + P(x>4.15)). Now we find those separated and, then, add them for our answer.
Let´s begin with P(x<3.85), we start by normalizing that probability:
P(x<3.85)= P( (x-μ)/σ < (3.85-4)/0.1 )= P(z<-1.5)
And now it´s time to use the chart, it works like this: If you want P(z<c) and the decimal expansion of c=a.bd... , then:
P(z<c)=(a.b , d)
Where (a.b , d) are the coordinates of the probability in the chart. Keep in mind that will only work with "<" (It won´t work directly with P(z>c)) and we will do some extra work in those cases.
P(z<-1.5) is in the coordinates (-1.5 , 0)
P(z<-1.5)= 0.0668
P(x<3.85)= 0.0668
Now we are looking for P(x>4.15), let´s Normalize it too:
P(x>4.15)=P( (x-μ)/σ < (4.15-4)/0.1 )=P(z>1.5)
But remember the chart only work with "<", so we need to use a property of probability:
P(z>1.5)= 1 - P(z<1.5)
Using the chart:
P(z<1.5)=0.9332 (1.5 , 0)
P(z>1.5)= 1 - 0.9332
P(z>1.5)= 0.0668
P(x>4.15)= 0.0668
And our final answer will be:
P(x<3.85 or x>4.15)= P(x<3.85)+P(x>4.15) = 0.1336
Complete question:
Consider a class of 20 students consisting of 5 sophomores, 8 juniors, and 7 seniors. A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from this class, keeping track (in order) of the level of the student that he gets.
1a. How many possible outcomes are in the sample space S? (An outcome is a 5- tuple of students.)
Let A2 denote the event that exactly 2 of the selected students are sophomore, A5 denote the event that exactly 5 of the selected students are the juniors, and A4 denote the event that exactly 4 of the selected students are seniors.
1b. Find the probabilities of each of these three events. A2, A5, A4
1c. Do the events A2, A5, A4 constitute a partition of the sample space?
Answer:
Given:
n = 20
a) The possible outcome in the sample space,S, =
ⁿCₓ = ²⁰C₅
= 15504
b) probabilities of A2, A5, A4
P(A2) = ⁵C₂ / ²⁰C₂
P(A5) = ⁸C₅ / ²⁰C₅
P(A4) = ⁷C₄ / ²⁰C₄
c) No,the events A2, A5, A4, do not comstitute a partition of the sample space. i.e P(A2)+P(A5)+P(A4) ≠ 1