91影视

The number of peas in one sample of offspring is recorded.

The given sample size$n=929$ and probability of success $p=0.75\left(\frac{3}{4}\right)$.

Then,

$$\begin{gathered} q=1-p \\ =1-0.75 \\ =0.25 \end{gathered}$$

From the given information,

$$\begin{gathered} np=929脳0.75 \\ =676.75 \\ >5 \end{gathered}$$

And

$$\begin{gathered} nq=929脳0.25 \\ =232.25 \\ >5 \end{gathered}$$

Here both and are greater than 5. Hence probabilities from a binomial

probability distribution can be approximated reasonably well by using a normal distribution.

The mean value is,

$$\begin{gathered} 渭=np \\ =929脳0.75 \\ =696.75 \end{gathered}$$

The standard deviation is,

$$\begin{gathered} 蟽=\sqrt{npq} \\ =\sqrt{929脳0.75脳0.25} \\ =13.20 \end{gathered}$$

a.

The probability of getting 705 or more peas with red flowers is expressed as using$P\left(X猢�n\right)\text{use}P\left(X>n-0.5\right)$

Thus

$$\begin{gathered} P\left(X猢�705\right)=P\left(X>705-0.5\right) \\ =P\left(X>704.5\right) \end{gathered}$$

Thus, the probability is expressed as $P\left(X>704.5\right)$.

The z-score for$x=704.5$,$渭=696.75,蟽=13.20$ is as follows:

$$\begin{gathered} z=\frac{x-渭}{蟽} \\ =\frac{704.5-696.75}{13.20} \\ =0.59 \end{gathered}$$

The z-score is 0.59.

Using the standard normal table, the cumulative probabilities are obtained for 0.59 as, 0.7224.

Substituting the value,

$$\begin{gathered} P\left(Z>0.59\right)=1-0.7224 \\ =0.2776 \end{gathered}$$

Thus, the probability is 0.2776.

b.

The event of getting 705 peas with red flowers would be significantly high if the probability of getting value 705 or above is lesser than 0.05.

In this case, 0.2776 > 0.05.

Therefore, it is not significantly high.

c.

As the value is not significantly high, it can be inferred that the chances of getting 705 or more peas do not suggest strong evidence against the claim that 0.75 proportion of peas will have red flowers.