Question

When a scientist conducted a genetics experiments with peas, one sample of offspring consisted of 943 peas, with 717 of them having red flowers. If we assume, as
the scientist did, that under these circumstances, there is a 3/4 probability that a pea will have a red flower, we would expect that 707.25 (or about 707) of the peas
would have red flowers, so the result of 717 peas with red flowers is more than expected.
a. If the scientist's assumed probability is correct, find the probability of getting 717 or more peas with red flowers
b. Is 717 peas with red flowers significantly high?
c. What do these results suggest about the scientist's assumption that 3/4 of peas will have red flowers?

Answer 1

Using the normal approximation to the binomial distribution, it is found that:

a) 0.242 = 24.2% probability of getting 717 or more peas with red flowers.

b) Since Z < 2, 717 peas with red flowers is not significantly high.

c) Since 717 peas with red flowers is not a significantly high result, we cannot conclude that the scientist's assumption is wrong.

For each pea, there are only two possible outcomes. Either they have a red flower, or they do not. The probability of a pea having a red flower is independent of any other pea, which means that the binomial distribution is used to solve this question.

Binomial distribution:

Probability of x successes on n trials, with p probability.

Normal distribution:

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  
• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

• If Z > 2, the result is considered significantly high.

If $np \geq 10$ and $n(1-p) \geq 10$, the binomial distribution can be approximated to the normal with:

$\mu = np$

$\sigma = \sqrt{np(1-p)}$

In this problem:

• 943 peas, thus, $n = 943$
• 3/4 probability of being red, thus $p = \frac{3}{4} = 0.75$.

Applying the approximation:

$\mu = np = 943(0.75) = 707.25$

$\sigma = \sqrt{np(1-p)} = \sqrt{943(0.75)(0.25)} = 13.297$

Item a:

Using continuity correction, this probability is $P(X \geq 717 - 0.5) = P(X \geq 716.5)$, which is 1 subtracted by the p-value of Z when X = 716.5.

Then:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{716.5 - 707.25}{13.297}$

$Z = 0.7$

$Z = 0.7$ has a p-value of 0.758.

1 - 0.758 = 0.242

0.242 = 24.2% probability of getting 717 or more peas with red flowers.

Item b:

Since Z < 2, 717 peas with red flowers is not significantly high.

Item c:

Since 717 peas with red flowers is not a significantly high result, we cannot conclude that the scientist's assumption is wrong.

A similar problem is given at brainly.com/question/25212369