Question

A researcher plants 22 seedlings. After one month, independent of the other seedlings, each seedling has a probability of 0.08 of being dead, a probability of 0.19 of exhibiting slow growth, a probability of 0.42 of exhibiting medium growth, and a probability of 0.31 of exhibiting strong growth. What is the expected number of seedlings in each of these four categories after one month? Calculate the probability that after one month:(a) Exactly three seedlings are dead, exactly four exhibitslow growth, and exactly six exhibit medium growth.(b) Exactly five seedlings are dead, exactly five exhibitslow growth, and exactly seven exhibit strong growth.(c) No more than two seedlings have died.

Answer 1

Answer:

E(X₁)= 1.76

E(X₂)= 4.18

E(X₃)= 9.24

E(X₄)= 6.82

a. P(X₁=3, X₂=4, X₃=6;0.08,0.19,0.42)= 0.00022

b. P(X₁=5, X₂=5, X₄=7;0.08,0.19,0.31)= 0.000001

c. P(X₁≤2) = 0.7442

Step-by-step explanation:

Hello!

So that you can easily resolve this problem first determine your experiment and it's variables. In this case, you have 22 seedlings (n) planted and observe what happens with the after one month, each seedling independent of the others and has each leads to success for exactly one of four categories with a fixed success probability per category. This is a multinomial experiment so I'll separate them in 4 different variables with the corresponding probability of success for each one of them:

X₁: "The seedling is dead" p₁: 0.08

X₂: "The seedling exhibits slow growth" p₂: 0.19

X₃: "The seedling exhibits medium growth" p₃: 0.42

X₄: "The seedling exhibits strong growth" p₄:0.31

To calculate the expected number for each category (k) you need to use the formula:

E(X$E(X_{k}) = n_{k} * p_{k}$

So

E(X₁)= n*p₁ = 22*0.08 = 1.76

E(X₂)= n*p₂ = 22*0.19 = 4.18

E(X₃)= n*p₃ = 22*0.42 = 9.24

E(X₄)= n*p₄ = 22*0.31 = 6.82

Next, to calculate each probability you just use the corresponding probability of success of each category:

Formula: P(X₁, X₂,..., Xk) = $\frac{n!}{X_{1}!X_{2}!...X_{k}!} * p_{1}^{X_{1}} * p_{2}^{X_{2}} *.....*p_{k}^{X_{k}}$

a.

P(X₁=3, X₂=4, X₃=6;0.08,0.19,0.42)= $\frac{22!}{3!4!6!} * 0.08^{3} * 0.19^{4} * 0.42^{6}$ = 0.00022

b.

P(X₁=5, X₂=5, X₄=7;0.08,0.19,0.31)= $\frac{22!}{5!5!7!} * 0.08^{5} * 0.19^{5} * 0.31^{7}$ = 0.000001

c.

P(X₁≤2) = $\frac{22!}{0!} * 0.08^{0} * (0.92)^{22}$ + $\frac{22!}{1!} * 0.08^{1} * (0.92)^{21}$ + $\frac{22!}{2!} * 0.08^{2} * (0.92)^{20}$ = 0.7442

I hope you have a SUPER day!