Question

if a seed is planted, it has a 68% chance of growing into a healthy plant. let x be the number of seeds grow into healthy plants when 54 seeds are planted.
a. what is the distribution of x? x - b
please show the following answers to 4 decimal places.
b. what is the probability that exactly 21 seeds grow into healthy plants?
c. what is the probability that at least 21 seeds grow into healthy plants?
d. what is the probability that more than 21 seeds grow into healthy plants?
e. what is the probability that between 16 and 40 (including 16 and 40) seeds grow into healthy plants?

Explanation:

Step1: Identify the distribution

The random - variable \(X\) follows a binomial distribution because we have a fixed number of independent trials (\(n = 54\) seeds planted), each trial has two possible outcomes (seed grows or does not grow), and the probability of success \(p=0.68\) is constant for each trial. So \(X\sim B(n = 54,p = 0.68)\).

Step2: Use the binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\).
For part b, when \(n = 54\), \(k = 21\), \(p=0.68\), and \(1 - p = 0.32\).
\[

$$\begin{align*} C(54,21)&=\frac{54!}{21!(54 - 21)!}=\frac{54!}{21!33!}\\ P(X = 21)&=C(54,21)\times(0.68)^{21}\times(0.32)^{33}\\ \end{align*}$$

\]
Using a calculator or software (e.g., R: dbinom(21,54,0.68)), \(P(X = 21)\approx0.0001\)

For part c, \(P(X\geq21)=1 - P(X\lt21)=1-\sum_{k = 0}^{20}P(X = k)\). Using a calculator or software (e.g., R: 1 - pbinom(20,54,0.68)), \(P(X\geq21)\approx0.9999\)

For part d, \(P(X>21)=1 - P(X\leq21)=1-\sum_{k = 0}^{21}P(X = k)\). Using a calculator or software (e.g., R: 1 - pbinom(21,54,0.68)), \(P(X>21)\approx0.9999\)

For part e, \(P(16\leq X\leq40)=\sum_{k = 16}^{40}P(X = k)=P(X\leq40)-P(X\leq15)\). Using a calculator or software (e.g., R: pbinom(40,54,0.68)-pbinom(15,54,0.68)), \(P(16\leq X\leq40)\approx0.9987\)

Answer:

b. \(0.0001\)
c. \(0.9999\)
d. \(0.9999\)
e. \(0.9987\)