91Ó°ÊÓ

A sample of 1064 offspring is considered, out of which 787 plants have long stems. The proportion of offspring plants with long stems is given to be equal to$\frac{3}{4}$

Let X (number of successes) denote the number of offspring plants with long stems.

The probability of success is given to be equal to:

$$\begin{gathered} p=\frac{3}{4} \\ =0.75 \end{gathered}$$

The number of trials (n) is equal to 1064.

Here, the sample is a result of 1064 independent trials with a probability of success at each trial equal to 0.75.

Also,

$$\begin{gathered} np=1064×0.75 \\ =798 \\ >5 \end{gathered}$$

$$\begin{gathered} nq=n(1-p) \\ =1064×(1-0.75) \\ =266 \\ >5 \end{gathered}$$

Since the above two requirements are met, the normal distribution can be used for approximating the binomial distribution.

It is required to compute the probability of 787 or fewer offspring plants with long stems.

Thus, the interval of continuity correction is computed below:

$$\begin{gathered} \left(x-0.5,x+0.5\right)=\left(787-0.5,787+0.5\right) \\ =\left(786.5,787.5\right) \end{gathered}$$

In terms of the bound of the continuity correction interval, the following probability needs to be computed:

$P\left(x<\text{upper}\text{bound}\right)=P\left(x<787.5\right)$

The sample value equal to x=787.5 is converted to a z-score as follows:

$$\begin{gathered} z=\frac{x-np}{\sqrt{np(1-p)}} \\ =\frac{787.5-1064(0.75)}{\sqrt{1064(0.75)(1-0.75)}} \\ =-0.74 \end{gathered}$$

Using the standard normal table, the probability of 787 or fewer plants with long stems is equal to:

$$\begin{gathered} P\left(x<787.5\right)=P\left(z<-0.74\right) \\ =0.2296 \end{gathered}$$

Thus, the probability of 787 or fewer offspring plants with long stems is approximately equal to 0.2296.

It can be said that the result of 787 plants with long stems is not significantly low as the probability value is high (not less than 0.05).

This suggests that there is insufficient evidence suggesting that Mendel’s proposed proportion of plants with long stems equal to 0.75 is incorrect. Thus, the test results support Mendel’s claimed proportion of $\frac{3}{4}$.