Math  /  Data & Statistics

QuestionAssume that hybridization experiments are conducted with peas having the property that for offspring, there is a 0.75 probability that a pea has green pods. Assume that the offspring peas are randomly selected in groups of 40 . Complete parts (a) through (c) below. a. Find the mean and the standard deviation for the numbers of peas with green pods in the groups of 40 .
The value of the mean is μ=\mu= \square peas. (Type an integer or a decimal. Do not round.)

Studdy Solution

STEP 1

What is this asking? We're looking at groups of 40 peas, and each pea has a 0.75 chance of having green pods.
We need to find the average number of green peas in these groups and how much that number typically varies. Watch out! Don't mix up the probability of a *single* pea being green with the average number of *green peas* in a group.
Also, remember that the standard deviation isn't the same as the variance!

STEP 2

1. Calculate the mean.
2. Calculate the standard deviation.

STEP 3

The **mean** number of peas with green pods is found by multiplying the number of peas in the group by the probability of a single pea having green pods.
It's like saying, "On average, how many peas do we *expect* to be green?"

STEP 4

We have 4040 peas in a group, and each has a 0.750.75 probability of being green.
So, the mean (μ\mu) is: μ=400.75=30 \mu = 40 \cdot 0.75 = \textbf{30} So, on average, we expect to see **30** green peas in a group of 40.

STEP 5

The **standard deviation** tells us how spread out the number of green peas is likely to be.
A small standard deviation means the numbers will usually be close to the mean, while a large standard deviation means they could be more spread out.

STEP 6

The formula for the standard deviation (σ\sigma) in a binomial distribution (like this pea situation) is: σ=np(1p) \sigma = \sqrt{n \cdot p \cdot (1 - p)} Where nn is the number of trials (peas in our case), and pp is the probability of success (a green pea).

STEP 7

Let's plug in our values. n=40n = 40 and p=0.75p = 0.75.
So, 1p=10.75=0.251 - p = 1 - 0.75 = 0.25. σ=400.750.25 \sigma = \sqrt{40 \cdot 0.75 \cdot 0.25}

STEP 8

Now, let's calculate inside the square root first: 400.750.25=7.5 40 \cdot 0.75 \cdot 0.25 = 7.5

STEP 9

So, we have: σ=7.52.739 \sigma = \sqrt{7.5} \approx \textbf{2.739} This tells us that the typical spread around our average of 30 green peas is about **2.739** peas.

STEP 10

The mean number of peas with green pods is **30**.
The standard deviation is approximately **2.739**.

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