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 $\mu=30$ peas. (Type an integer or a decimal. Do not round.) The value of the standard deviation is $\sigma=\square$ $\square$ peas. (Round to one decimal place as needed.)

Studdy Solution

STEP 1

What is this asking? We're looking at groups of 40 peas, where each pea has a 0.75 chance of having green pods.
We need to find the average number of green-pod peas in these groups and how much this number typically varies. Watch out! Don't mix up the probability (0.75) with the number of peas (40).
Also, remember to round the standard deviation correctly!

STEP 2

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

STEP 3

We're given that each pea has a $\text{0.75}$ probability of having green pods, and we have a group of $\text{40}$ peas.
The mean, represented by $\mu$ , is the average number of green-pod peas we expect to find in a group.

STEP 4

To calculate the mean, we multiply the number of peas by the probability of a pea having green pods: $\mu = n \cdot p$ Where $n$ is the number of peas ( $n = 40$ ) and $p$ is the probability of a pea having green pods ( $p = 0.75$ ).

STEP 5

Let's plug in the values: $\mu = 40 \cdot 0.75 = \textbf{30}$ So, on average, we expect to see 30 peas with green pods in a group of 40.

STEP 6

The standard deviation, represented by $\sigma$ , tells us how much the actual number of green-pod peas is likely to vary from the mean.

STEP 7

The formula for the standard deviation of a binomial distribution (like this one) is: $\sigma = \sqrt{n \cdot p \cdot (1 - p)}$ Where $n$ is the number of peas ( $n = 40$ ), and $p$ is the probability of a pea having green pods ( $p = 0.75$ ).
Notice that $(1 - p)$ is the probability of a pea not having green pods.

STEP 8

Let's plug in the values: $\sigma = \sqrt{40 \cdot 0.75 \cdot (1 - 0.75)}$ $\sigma = \sqrt{40 \cdot 0.75 \cdot 0.25}$ $\sigma = \sqrt{7.5}$ $\sigma \approx \textbf{2.7386}$

STEP 9

We need to round the standard deviation to one decimal place, so the final result is approximately $\textbf{2.7}$ .

STEP 10

The mean number of peas with green pods is $\textbf{30}$ .
The standard deviation is approximately $\textbf{2.7}$ .

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Studdy Solution

STEP 1

STEP 2

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

STEP 3

We're given that each pea has a $\text{0.75}$ probability of having green pods, and we have a group of $\text{40}$ peas.
The mean, represented by $\mu$ , is the average number of green-pod peas we expect to find in a group.

STEP 4

To calculate the mean, we multiply the number of peas by the probability of a pea having green pods: $\mu = n \cdot p$ Where $n$ is the number of peas ( $n = 40$ ) and $p$ is the probability of a pea having green pods ( $p = 0.75$ ).

STEP 5

Let's plug in the values: $\mu = 40 \cdot 0.75 = \textbf{30}$ So, on average, we expect to see 30 peas with green pods in a group of 40.

STEP 6

The standard deviation, represented by $\sigma$ , tells us how much the actual number of green-pod peas is likely to vary from the mean.

STEP 7

STEP 8

Let's plug in the values: $\sigma = \sqrt{40 \cdot 0.75 \cdot (1 - 0.75)}$ $\sigma = \sqrt{40 \cdot 0.75 \cdot 0.25}$ $\sigma = \sqrt{7.5}$ $\sigma \approx \textbf{2.7386}$

STEP 9

We need to round the standard deviation to one decimal place, so the final result is approximately $\textbf{2.7}$ .

STEP 10

The mean number of peas with green pods is $\textbf{30}$ .
The standard deviation is approximately $\textbf{2.7}$ .

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Studdy Solution

STEP 1

STEP 2

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

STEP 3

We're given that each pea has a 0.75\text{0.75}0.75 probability of having green pods, and we have a group of 40\text{40}40 peas.The mean, represented by μ\muμ, is the average number of green-pod peas we expect to find in a group.

STEP 4

To **calculate the mean**, we multiply the number of peas by the probability of a pea having green pods: μ=n⋅p \mu = n \cdot p μ=n⋅p Where nnn is the number of peas (n=40n = 40n=40) and ppp is the probability of a pea having green pods (p=0.75p = 0.75p=0.75).

STEP 5

Let's plug in the values: μ=40⋅0.75=30 \mu = 40 \cdot 0.75 = \textbf{30} μ=40⋅0.75=30 So, on average, we expect to see **30** peas with green pods in a group of 40.

STEP 6

The standard deviation, represented by σ\sigmaσ, tells us how much the actual number of green-pod peas is likely to vary from the mean.

STEP 7

STEP 8

STEP 9

We need to round the standard deviation to one decimal place, so the **final result** is approximately 2.7\textbf{2.7}2.7.

STEP 10

The mean number of peas with green pods is 30\textbf{30}30.The standard deviation is approximately 2.7\textbf{2.7}2.7.

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

We're given that each pea has a $\text{0.75}$ probability of having green pods, and we have a group of $\text{40}$ peas.
The mean, represented by $\mu$ , is the average number of green-pod peas we expect to find in a group.

To calculate the mean, we multiply the number of peas by the probability of a pea having green pods: $\mu = n \cdot p$ Where $n$ is the number of peas ( $n = 40$ ) and $p$ is the probability of a pea having green pods ( $p = 0.75$ ).

Let's plug in the values: $\mu = 40 \cdot 0.75 = \textbf{30}$ So, on average, we expect to see 30 peas with green pods in a group of 40.

The standard deviation, represented by $\sigma$ , tells us how much the actual number of green-pod peas is likely to vary from the mean.

We need to round the standard deviation to one decimal place, so the final result is approximately $\textbf{2.7}$ .

The mean number of peas with green pods is $\textbf{30}$ .
The standard deviation is approximately $\textbf{2.7}$ .