Question

if a seed is planted, it has a 37% chance of growing into a healthy plant. let x be the number of seeds grow into healthy plants when 50 seeds are planted.
a. what is the distribution of x? x ~?
please show the following answers to 4 decimal places.
b. what is the probability that exactly 24 seeds grow into healthy plants?
c. what is the probability that at most 24 seeds grow into healthy plants?
d. what is the probability that at least 24 seeds grow into healthy plants?
e. what is the probability that between 22 and 32 (including 22 and 32) 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 (\(n = 50\) seed - plantings), each with two possible outcomes (seed grows or does not grow), and a constant probability of success (\(p=0.37\)). So \(X\sim B(n = 50,p = 0.37)\).

Step2: Calculate the probability mass function for part b

The probability mass function of a binomial distribution 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 \(n = 50\), \(k = 24\), and \(p = 0.37\), we first calculate \(C(50,24)=\frac{50!}{24!(50 - 24)!}=\frac{50!}{24!26!}\), then \(P(X = 24)=\frac{50!}{24!26!}\times(0.37)^{24}\times(0.63)^{26}\).
\[C(50,24)=\frac{50!}{24!(50 - 24)!}=\frac{50\times49\times\cdots\times27}{24!}\approx1.477\times10^{13}\]
\[P(X = 24)\approx1.477\times10^{13}\times(0.37)^{24}\times(0.63)^{26}\approx0.0533\]

Step3: Calculate the cumulative - distribution function for part c

The cumulative - distribution function of a binomial distribution \(B(n,p)\) is \(P(X\leq k)=\sum_{i = 0}^{k}C(n,i)\times p^{i}\times(1 - p)^{n - i}\). We can use a statistical software or a calculator with binomial - cumulative distribution function. Using a calculator, \(P(X\leq24)=\sum_{i = 0}^{24}\frac{50!}{i!(50 - i)!}\times(0.37)^{i}\times(0.63)^{50 - i}\approx0.9131\).

Step4: Calculate the probability for part d

\(P(X\geq24)=1 - P(X\leq23)\). First, find \(P(X\leq23)=\sum_{i = 0}^{23}\frac{50!}{i!(50 - i)!}\times(0.37)^{i}\times(0.63)^{50 - i}\approx0.8598\). Then \(P(X\geq24)=1 - 0.8598 = 0.1402\).

Step5: Calculate the probability for part e

\(P(22\leq X\leq32)=\sum_{i = 22}^{32}\frac{50!}{i!(50 - i)!}\times(0.37)^{i}\times(0.63)^{50 - i}\). Using a calculator or software, \(P(22\leq X\leq32)\approx0.7739\).

Answer:

a. \(X\sim B(50,0.37)\)
b. \(0.0533\)
c. \(0.9131\)
d. \(0.1402\)
e. \(0.7739\)