QuestionFind probabilities for lemonade consumption: a) > 21 gallons, b) < 19 gallons, c) between 20 and 25 gallons. For sodium: a) > 670 mg in one dinner, c) mean > 670 mg in 10 dinners. Also, estimate the mean LDL cholesterol change from garlic treatment.

Studdy Solution

STEP 1

Assumptions1. The average number of gallons of lemonade consumed by the football team during a game is20. The standard deviation is3 gallons3. The variable is normally distributed

STEP 2

We can use the z-score formula to find the probability of using more than21 gallons. The z-score is calculated as follows $Z = (X - \mu) / \sigma$ where X is the value we are interested in, $\mu$ is the mean, and $\sigma$ is the standard deviation.

STEP 3

Plug in the values for X, $\mu$ , and $\sigma$ to calculate the z-score for21 gallons.
$Z = (21 -20) /3$

STEP 4

Calculate the z-score.
$Z = (21 -20) /3 =0.33$

STEP 5

The probability of using more than21 gallons is the area under the normal distribution curve to the right of the z-score. This is found by subtracting the cumulative probability at the z-score from1.
$(X >21) =1 -(Z <0.33)$

STEP 6

Using a standard normal distribution table or a calculator, we find that $(Z <0.33) \approx0.6293$ .

STEP 7

Subtract the cumulative probability at the z-score from1 to find the probability of using more than21 gallons.
$(X >21) =1 -0.6293 =0.3707$

STEP 8

Repeat steps2-7 for X =19 to find the probability of using less than19 gallons.

STEP 9

Repeat steps2-7 for X =20 and X =25 to find the probabilities of using these amounts. The probability of using between20 and25 gallons is the difference between these two probabilities.

STEP 10

For the sodium content problem, repeat the steps above, but with $\mu =660$ and $\sigma =35$ .

STEP 11

For the sample of10 dinners, use the standard error of the mean, which is $\sigma / \sqrt{n}$ , where n is the sample size. Then repeat the steps above.

STEP 12

For the cholesterol problem, the best point estimate of the mean is the sample mean, which is given as.2.

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QuestionFind probabilities for lemonade consumption: a) > 21 gallons, b) < 19 gallons, c) between 20 and 25 gallons. For sodium: a) > 670 mg in one dinner, c) mean > 670 mg in 10 dinners. Also, estimate the mean LDL cholesterol change from garlic treatment.

Studdy Solution

STEP 1

Assumptions1. The average number of gallons of lemonade consumed by the football team during a game is20. The standard deviation is3 gallons3. The variable is normally distributed

STEP 2

We can use the z-score formula to find the probability of using more than21 gallons. The z-score is calculated as follows $Z = (X - \mu) / \sigma$ where X is the value we are interested in, $\mu$ is the mean, and $\sigma$ is the standard deviation.

STEP 3

Plug in the values for X, $\mu$ , and $\sigma$ to calculate the z-score for21 gallons.
$Z = (21 -20) /3$

STEP 4

Calculate the z-score.
$Z = (21 -20) /3 =0.33$

STEP 5

The probability of using more than21 gallons is the area under the normal distribution curve to the right of the z-score. This is found by subtracting the cumulative probability at the z-score from1.
$(X >21) =1 -(Z <0.33)$

STEP 6

Using a standard normal distribution table or a calculator, we find that $(Z <0.33) \approx0.6293$ .

STEP 7

Subtract the cumulative probability at the z-score from1 to find the probability of using more than21 gallons.
$(X >21) =1 -0.6293 =0.3707$

STEP 8

Repeat steps2-7 for X =19 to find the probability of using less than19 gallons.

STEP 9

Repeat steps2-7 for X =20 and X =25 to find the probabilities of using these amounts. The probability of using between20 and25 gallons is the difference between these two probabilities.

STEP 10

For the sodium content problem, repeat the steps above, but with $\mu =660$ and $\sigma =35$ .

STEP 11

For the sample of10 dinners, use the standard error of the mean, which is $\sigma / \sqrt{n}$ , where n is the sample size. Then repeat the steps above.

STEP 12

For the cholesterol problem, the best point estimate of the mean is the sample mean, which is given as.2.

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QuestionFind probabilities for lemonade consumption: a) > 21 gallons, b) < 19 gallons, c) between 20 and 25 gallons. For sodium: a) > 670 mg in one dinner, c) mean > 670 mg in 10 dinners. Also, estimate the mean LDL cholesterol change from garlic treatment.

Studdy Solution

STEP 1

Assumptions1. The average number of gallons of lemonade consumed by the football team during a game is20. The standard deviation is3 gallons3. The variable is normally distributed

STEP 2

We can use the z-score formula to find the probability of using more than21 gallons. The z-score is calculated as followsZ=(X−μ)/σZ = (X - \mu) / \sigmaZ=(X−μ)/σwhere X is the value we are interested in, μ\muμ is the mean, and σ\sigmaσ is the standard deviation.

STEP 3

Plug in the values for X, μ\muμ, and σ\sigmaσ to calculate the z-score for21 gallons.Z=(21−20)/3Z = (21 -20) /3Z=(21−20)/3

STEP 4

Calculate the z-score.Z=(21−20)/3=0.33Z = (21 -20) /3 =0.33Z=(21−20)/3=0.33

STEP 5

The probability of using more than21 gallons is the area under the normal distribution curve to the right of the z-score. This is found by subtracting the cumulative probability at the z-score from1.(X>21)=1−(Z<0.33)(X >21) =1 -(Z <0.33)(X>21)=1−(Z<0.33)

STEP 6

Using a standard normal distribution table or a calculator, we find that (Z<0.33)≈0.6293(Z <0.33) \approx0.6293(Z<0.33)≈0.6293.

STEP 7

Subtract the cumulative probability at the z-score from1 to find the probability of using more than21 gallons.(X>21)=1−0.6293=0.3707(X >21) =1 -0.6293 =0.3707(X>21)=1−0.6293=0.3707

STEP 8

Repeat steps2-7 for X =19 to find the probability of using less than19 gallons.

STEP 9

Repeat steps2-7 for X =20 and X =25 to find the probabilities of using these amounts. The probability of using between20 and25 gallons is the difference between these two probabilities.

STEP 10

For the sodium content problem, repeat the steps above, but with μ=660\mu =660μ=660 and σ=35\sigma =35σ=35.

STEP 11

For the sample of10 dinners, use the standard error of the mean, which is σ/n\sigma / \sqrt{n}σ/n​, where n is the sample size. Then repeat the steps above.

STEP 12

For the cholesterol problem, the best point estimate of the mean is the sample mean, which is given as.2.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

We can use the z-score formula to find the probability of using more than21 gallons. The z-score is calculated as follows $Z = (X - \mu) / \sigma$ where X is the value we are interested in, $\mu$ is the mean, and $\sigma$ is the standard deviation.

Plug in the values for X, $\mu$ , and $\sigma$ to calculate the z-score for21 gallons.
$Z = (21 -20) /3$

Calculate the z-score.
$Z = (21 -20) /3 =0.33$

The probability of using more than21 gallons is the area under the normal distribution curve to the right of the z-score. This is found by subtracting the cumulative probability at the z-score from1.
$(X >21) =1 -(Z <0.33)$

Using a standard normal distribution table or a calculator, we find that $(Z <0.33) \approx0.6293$ .

Subtract the cumulative probability at the z-score from1 to find the probability of using more than21 gallons.
$(X >21) =1 -0.6293 =0.3707$

For the sodium content problem, repeat the steps above, but with $\mu =660$ and $\sigma =35$ .

For the sample of10 dinners, use the standard error of the mean, which is $\sigma / \sqrt{n}$ , where n is the sample size. Then repeat the steps above.