Ultimate Binomial Exercises! Exercises 37–40 involve finding binomial probabilities, finding parameters, and determining whether values are significantly high or low by using the range rule of thumb and probabilities.

Hybrids One of Mendel’s famous experiments with peas resulted in 580 offspring, and 152 of them were yellow peas. Mendel claimed that under the same conditions, 25% of offspring peas would be yellow. Assume that Mendel’s claim of 25% is true, and assume that a sample consists of 580 offspring peas. a. Use the range rule of thumb to identify the limits separating values that are significantly low and those that are significantly high. Based on the results, is the result of 152 yellow peas either significantly low or significantly high?

b. Find the probability of exactly 152 yellow peas.

c. Find the probability of 152 or more yellow peas.

d. Which probability is relevant for determining whether 152 peas is significantly high: the probability from part (b) or part (c)? Based on the relevant probability, is the result of 152 yellow peas significantly high?

e. What do the results suggest about Mendel’s claim of 25%?

It is given that 25% of offspring peas should be yellow.

A total of 580 offspring of peas were selected, and 152 of them were yellow.

The total number of offspring of peas selected (n) is equal to 580.

The probability of selecting a yellow offspring is given as follows:

$$\begin{gathered} p=25\% \\ =\frac{25}{100} \\ =0.25 \end{gathered}$$

The mean of the number of offspring that are yellow is given as follows:

$$\begin{gathered} \mu =np \\ =580\left(0.25\right) \\ =145 \end{gathered}$$

Thus, $\mu =145$.

The standard deviation for the offspring peas to be yellow is computed below:

$$\begin{gathered} \sigma =\sqrt{npq} \\ =\sqrt{np\left(1-p\right)} \\ =\sqrt{580\times 0.25\times \left(1-0.25\right)} \\ =10.43 \end{gathered}$$

Thus, $\sigma =10.43$.

a.

The following limit separates significantly low values:

$$\begin{gathered} \mu -2\sigma =145-\left(2\right)\left(10.43\right) \\ =124.14 \end{gathered}$$

Therefore, values less than or equal to 124.14 are considered to be significantly low.

The following limit separates significantly high values:

$$\begin{gathered} \mu +2\sigma =145+\left(2\right)\left(10.43\right) \\ =165.86 \end{gathered}$$

Therefore, values greater than or equal to 165.86 are considered to be significantly high.

Values lying between124.14 and 165.86 are not significant.

Here, the value of 152 yellow offspring lies between 124.14 and 165.86.

Thus, it is neither significantly high nor significantly low.

b.

Let X denote the number of yellow offspring peas.

Success is defined as selecting a yellow offspring.

The probability of success is p=0.25.

The probability of failure is computed below:

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

The number of trials (n) is equal to 580.

The binomial probability formula used to compute the given probability is as follows:

$P\left(X=x\right)={}^n{C}_x{\left(p\right)}^x{\left(q\right)}^{n-x}$

Using the binomial probability formula, the probability of getting 152 yellow offspring peas is as follows:

$$\begin{gathered} P\left(X=152\right)={}^{580}{C}_{152}{\left(0.25\right)}^{152}{\left(0.75\right)}^{580-152} \\ =0.0301 \end{gathered}$$

Thus, the probability of getting 152 yellow offspring peas is equal to 0.0301.

c.

The probability of 152 or more yellow offspring peas has the following expression:

$$\begin{gathered} P\left(X⩾152\right)=1-P\left(X<152\right) \\ =1-\left[P\left(X=0\right)+P\left(X=1\right)+......+P\left(X=151\right)\right] \end{gathered}$$

The individual probabilities will be computed as follows:

$$\begin{gathered} P\left(X=0\right)={}^{580}{C}_0{\left(0.25\right)}^0{\left(0.75\right)}^{508} \\ \approx 3.432\times {10}^{-73} \\ P\left(X=1\right)={}^{580}{C}_1{\left(0.25\right)}^1{\left(0.75\right)}^{579} \\ \approx 6.635\times {10}^{-71} \\ . \\ . \\ . \\ P\left(X=151\right)={}^{580}{C}_{151}{\left(0.25\right)}^{151}{\left(0.75\right)}^{429} \\ \approx 0.032018979 \end{gathered}$$

Thus, the required probability is computed as follows:

$$\begin{gathered} P\left(X⩾152\right)=1-\left[P\left(X=0\right)+P\left(X=1\right)+......+P\left(X=151\right)\right] \\ =1-0.73504417 \\ \approx 0.265 \end{gathered}$$

Thus, the probability of 152 or more yellow offspring peas is equal to 0.265.

d.

The probability formula to conclude that the given value of the number of successes (x) is significantly high is shown below:

$P\left(x\text{or}\text{more}\right)⩽0.05$

Here, the value of x considered is equal to 152

Since the probability of152 or more yellow offspring peasis represented in part (c). Thus, it can be concluded that theprobability computed in part (c) is relevant for determining whether the result of 152 yellow offspring peas is significantly high.

Thus,

$$\begin{gathered} P\left(152\text{or}\text{more}\right)=0.265 \\ >0.05 \end{gathered}$$

Since the probability of 152 or more yellow offspring peas is not less than or equal to 0.05, the value of 152 yellow offspring peas is not significantly high.

e.

As per Mendel’s claim, 25% of the peas should be yellow.

Out of 580 offspring, the expected number of yellow peas, according to Mendel’s claim, should be equal to 145.

The given value of yellow offspring in the sample of 580 offspring is equal to 152.

As the given value and the expected value are quite close, it can be concluded that Mendel’s claim of 25% yellow offspring peas is valid.