Question

79. Sowing seeds Refer to Exercise 69.

(a) Find the probability that exactly$17$ seeds germinate. Show your work.
(b) If only $12$seeds actually germinate, should Judy be suspicious that the company’s claim is not true? Compute $P(X\le 12)$and use this result to support your answer.

Short answer

(a) The probability that exactly $17$seeds germinate is $\mathrm{P}(\mathrm{X}=17)=24.28\%$.

(b) Judy should be suspicious and the value is$P(X\le 12)\approx 0.0059\approx 0.59\%$.

Step-by-step solution

Part (a) Step 1: Given information 

The probability for the exactly $17$seeds germinate.

Part (a) Step 2: Explanation 

Given: $n=20$, and $p=0.85$
The Binomial Probability:
$\mathrm{P}(\mathrm{X}=\mathrm{k})=\left(\begin{array}{c}n\\ k\end{array}\right)\times p^k\times (1-p{)}^{n-k}$

$$\begin{gathered} \mathrm{P}(\mathrm{X}=17)=\left(\begin{array}{c}20\\ 17\end{array}\right)\times (0.85{)}^{17}\times (0.15{)}^{20-17} \\ \approx 0.2428 \\ =24.28\mathrm{\%} \end{gathered}$$

Part (b) Step 1: Given information 

Let,$12$ seeds actually germinate. And compute the value $P(X\le 12)$.

Part (b) Step 2: Explanation 

Given: $n=20$, and $p=0.85$.

Let's use the binomial Probability:

$\mathrm{P}(\mathrm{X}=\mathrm{k})=\left(\begin{array}{l}n\\ k\end{array}\right)\times p^k\times (1-p{)}^{n-k}$
$$\begin{gathered} \mathrm{P}(\mathrm{X}\le 12)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\dots +\mathrm{P}(\mathrm{X}=12) \\ \approx 0.0059 \\ \approx 0.59\mathrm{\%} \end{gathered}$$

As a result, the probability is low (less than$0.05$), indicating that the event is unlikely to happen by chance.