Mathematics: On a standardized test with normal distribution, the mean is 85 and the standard deviation is 3. Within which range should approximately 95% of the score lie?

On a standardized test with normal distribution,, №15171902, 04.02.2023 10:42

Mathematics

The 68-95-99.7 Rule About 68% of the data points fall within 1 standard deviation of the mean About 95% of the data points fall within 2 standard deviation of the mean About 99.7% of the data points fall within 3 standard deviations of the mean Assume that a set test scores is normally distributed with a mean of 100 and a standard deviation of 20. Use the 68-95-99.7 rule to find the following 1) Percentages of scores less than 100 2) Percentage of scores less than 140 3) Percentage of scores less than 80 4) Percentage of scores between 80 and 120

Step-by-step answer

P Answered by Master

1) Percentages of scores less than 100: 50%

2) Percentage of scores less than 140: 97.5%

3) Percentage of scores less than 80: 16%

4) Percentage of scores between 80 and 120: 68%

5) Percentage of scores between 80 and 140: 81.5%

1) Percentage of rates less than 70 : 50%

2) Percentage of rates less than 55 : 16%

3) Percentage of rates less than 85 : 97.5%

4) Percentage of rates greater than 85 : 2.5%

5) Percentage of rates greater than 55 : 84%

6) Percentage of rates between 55 and 100: 81.5%

7) Percentage of rates between 70 and 100: 47.5%

Step-by-step explanation:

We have a random variable normally distributed with a mean of 100 and a standard deviation of 20.

1) Percentages of scores less than 100: 50%

As the mean is 100, 50% of the data lies below 100.

2) Percentage of scores less than 140: 97.5%

The data is what lies below (mean +2 sd). In this case, applies the 95% rule for the higher scores (above 100), which means we have 95/2=47.5 of the data between 100 and 140.

The data below 100 represents 50%.

So the scores under 140 are 50+47.5=97.5%.

3) Percentage of scores less than 80: 16%

The scores under 80 are a (mean-1 sd).

This means that is half of the data, less 68/2=34 (the area that is under the first standard deviation of the mean and the mean).

Then, the scores under 80 are 50-34=16%.

4) Percentage of scores between 80 and 120: 68%

The scores are under one deviation of the mean (to both sides). The 68% rule applies.

5) Percentage of scores between 80 and 140: 81.5%

The lower scores, between 80 and 100 are in the area between one deviation and the mean, so it has a percentage of 68/2=34%.

The higher scores are 2 deviations frome the mean, so they have 95/2=47.5% of the scores.

Between 80 and 140 are 34+47.5=81.5% of the scores.

Mathematics

The 68-95-99.7 Rule About 68% of the data points fall within 1 standard deviation of the mean About 95% of the data points fall within 2 standard deviation of the mean About 99.7% of the data points fall within 3 standard deviations of the mean Assume that a set test scores is normally distributed with a mean of 100 and a standard deviation of 20. Use the 68-95-99.7 rule to find the following 1) Percentages of scores less than 100 2) Percentage of scores less than 140 3) Percentage of scores less than 80 4) Percentage of scores between 80 and 120

Step-by-step answer

P Answered by Specialist

1) Percentages of scores less than 100: 50%

2) Percentage of scores less than 140: 97.5%

3) Percentage of scores less than 80: 16%

4) Percentage of scores between 80 and 120: 68%

5) Percentage of scores between 80 and 140: 81.5%

1) Percentage of rates less than 70 : 50%

2) Percentage of rates less than 55 : 16%

3) Percentage of rates less than 85 : 97.5%

4) Percentage of rates greater than 85 : 2.5%

5) Percentage of rates greater than 55 : 84%

6) Percentage of rates between 55 and 100: 81.5%

7) Percentage of rates between 70 and 100: 47.5%

Step-by-step explanation:

We have a random variable normally distributed with a mean of 100 and a standard deviation of 20.

1) Percentages of scores less than 100: 50%

As the mean is 100, 50% of the data lies below 100.

2) Percentage of scores less than 140: 97.5%

The data is what lies below (mean +2 sd). In this case, applies the 95% rule for the higher scores (above 100), which means we have 95/2=47.5 of the data between 100 and 140.

The data below 100 represents 50%.

So the scores under 140 are 50+47.5=97.5%.

3) Percentage of scores less than 80: 16%

The scores under 80 are a (mean-1 sd).

This means that is half of the data, less 68/2=34 (the area that is under the first standard deviation of the mean and the mean).

Then, the scores under 80 are 50-34=16%.

4) Percentage of scores between 80 and 120: 68%

The scores are under one deviation of the mean (to both sides). The 68% rule applies.

5) Percentage of scores between 80 and 140: 81.5%

The lower scores, between 80 and 100 are in the area between one deviation and the mean, so it has a percentage of 68/2=34%.

The higher scores are 2 deviations frome the mean, so they have 95/2=47.5% of the scores.

Between 80 and 140 are 34+47.5=81.5% of the scores.

Mathematics

A researcher wants to investigate claims that a new plant compound is effective in curbing appetite. She randomly assigns 26 rats to a treatment or control group (13 rats in each group). Rats in the treatment group are given the plant compound daily for a week, while rats in the control group are given a placebo. Over the course of the week, the researcher monitors the total number of calories consumed by all 26 rats. All the rats are kept separate from one another, but in their typical environments, for the duration of the week. The researcher’s

Step-by-step answer

P Answered by Master

Answer:

Yes, you don't use a pooled variance and you recompute the degrees of freedom so that the resulting df is smaller.

Step-by-step explanation:

Yes, you don't use a pooled variance and you recompute the degrees of freedom so that the resulting df is smaller.

Mathematics

Suppose a horticulturist measures the aboveground height growth rate of four different ornamental shrub species grown in a greenhouse. The shrubs were grown from a random sample of seeds, and they were all grown in the same soil mixture and in the same size pot. To ensure that any slight differences in the environmental conditions throughout the greenhouse are not confounded with species, she randomizes the location of the pots throughout the greenhouse. The table contains a summary of her data. Population Sample Sample Sample Population description

Step-by-step answer

P Answered by Specialist

Complete Question

The complete question is shown on the first uploaded image

The decision is to reject the null hypothesis at a significant level of significance $\alpha = 0.05$ There is sufficient evidence to conclude that at least one of the population mean is different from at least of the population

Step-by-step explanation:

From the question we are told that the claim is

The mean growth rates of all four species are equal.

The null hypothesis is

$H_o : \mu _1 = \mu_2 = \mu_3 = \mu_4$

Th alternative hypothesis is

$H_a: at \ least \ one \ of \ the \ means \ is \not\ equal$

From question the p-value is p-value = 0.015

And since the $p-value < \alpha$ so the null hypothesis will be rejected

So

The decision is to reject the null hypothesis at a significant level of significance $\alpha = 0.05$ There is sufficient evidence to conclude that at least one of the population mean is different from at least of the population

Suppose a horticulturist measures the aboveground height growth rate of four different ornamental sh

Mathematics

Suppose a horticulturist measures the aboveground height growth rate of four different ornamental shrub species grown in a greenhouse. The shrubs were grown from a random sample of seeds, and they were all grown in the same soil mixture and in the same size pot. To ensure that any slight differences in the environmental conditions throughout the greenhouse are not confounded with species, she randomizes the location of the pots throughout the greenhouse. The table contains a summary of her data. Population Sample Sample Sample Population description

Step-by-step answer

P Answered by Specialist

Complete Question

The complete question is shown on the first uploaded image

The decision is to reject the null hypothesis at a significant level of significance $\alpha = 0.05$ There is sufficient evidence to conclude that at least one of the population mean is different from at least of the population

Step-by-step explanation:

From the question we are told that the claim is

The mean growth rates of all four species are equal.

The null hypothesis is

$H_o : \mu _1 = \mu_2 = \mu_3 = \mu_4$

Th alternative hypothesis is

$H_a: at \ least \ one \ of \ the \ means \ is \not\ equal$

From question the p-value is p-value = 0.015

And since the $p-value < \alpha$ so the null hypothesis will be rejected

So

The decision is to reject the null hypothesis at a significant level of significance $\alpha = 0.05$ There is sufficient evidence to conclude that at least one of the population mean is different from at least of the population

Mathematics

Imran is paid ?286 per week Every week he puts 20% of his wage into a savings account How much does he put into his savings each week

Step-by-step answer

P Answered by PhD

The answer is in the image