Question

Hybridization Experiment In one of Mendel’s experiments with plants, 1064 offspring consisted of 787 plants with long stems. According to Mendel’s theory, 3/4 of the offspring plants should have long stems. Assuming that Mendel’s proportion of 3/4 is correct, find the probability of getting 787 or fewer plants with long stems among 1064 offspring plants. Based on the result, is 787 offspring plants with long stems significantly low? What does the result imply about Mendel’s claimed proportion of 3/4?

Short answer

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

The result of 787 plants with long stems is not significantly low as the probability value is high (not less than 0.05).

There is not enough evidence to suggest that the proportion of plants with long stems, as proposed by Mendel (0.75), is incorrect.

Step-by-step solution

Given information

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$\frac{3}{4}$

Requirements

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,

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

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

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

Continuity correction

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:

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

Conversion of sample value to z-score

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

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

Required probability

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.

Interpretation of the probability value

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 $\frac{3}{4}$.