A seed company has a test plot in which it is testing the germination of a hybrid seed. They plant 50 rows of 40seeds per row. After a two-week period, the researchers count how many seed per row have sprouted. Theynoted that least number of seeds to germinate was 33 and some rows had all 40 germinate. The germinationdata is given below in the table. The random variable x represents the number of seed in a row thatgerminated and P(x) represents the probability of selecting a row with that number of seed germinating.Determine the standard deviation of the number of seeds per row that germinated. $$\begin{array} { | c | c | c | c | c | c | c | c | c | }
\hline x & 33 & 34 & 35 & 36 & 37 & 38 & 39 & 40 \
\hline P ( x ) & 0.02 & 0.06 & 0.10 & 0.20 & 0.24 & 0.26 & 0.10 & 0.02 \
\hline
\end{array}$$
A) 1.51
B) 7.13
C) 36.86
D) 6.07 6.2 The Binomial Probability Distribution
1 Determine whether a probability experiment is a binomial experiment.

Question

A seed company has a test plot in which it is testing the germination of a hybrid seed. They plant 50 rows of 40seeds per row. After a two-week period, the researchers count how many seed per row have sprouted. Theynoted that least number of seeds to germinate was 33 and some rows had all 40 germinate. The germinationdata is given below in the table. The random variable x represents the number of seed in a row thatgerminated and P(x) represents the probability of selecting a row with that number of seed germinating.Determine the standard deviation of the number of seeds per row that germinated. $$\begin{array} { | c | c | c | c | c | c | c | c | c | } 
\hline x & 33 & 34 & 35 & 36 & 37 & 38 & 39 & 40 \
\hline P ( x ) & 0.02 & 0.06 & 0.10 & 0.20 & 0.24 & 0.26 & 0.10 & 0.02 \
\hline
\end{array}$$
A) 1.51
B) 7.13
C) 36.86
D) 6.07 6.2 The Binomial Probability Distribution
1 Determine whether a probability experiment is a binomial experiment.

Quizplus · Accepted Answer

The standard deviation is calculated using the formula for a discrete probability distribution: $\sigma = \sqrt{\sum (x - \mu)^2 P(x)}$, where $\mu$ is the mean. Calculating this for the given data results in a standard deviation of approximately 1.51.

A Seed Company Has a Test Plot in Which It