What is a standard probability distribution?

Marine Height Requirement for Women:

Marine Height Requirement for Women: To be eligible for the U.S. Marine Corps, a woman must have a height between 58 in and 73 in. Women have normally distributed heights with a mean of 63.6 in. and a standard deviation of 2.5 in.

1. Find the percentage of women who satisfy the Marine Corps height requirement. (5 points)
2. If the requirement is change to exclude the shortest \(1 \%\) and the tallest \(1 \%\), find the minimum and maximum heights that are acceptable. ( 5 points)
3. If one women are randomly selected, find the probability that their mean heights between 63.0 in. and 65.0 in. ( 5 points)
4. If nine women are randomly selected, find the probability that their mean heights between 63.0 in. and 65.0 in. ( 5 points)

MCAT Scores:

MCAT Scores: Scores on the biology portion of the Medical College Admissions Test are normally distributed with a mean of 9.1 and a standard deviation of 2.1.

1. What percentage of scores is above 8.5, the minimum score required for admission? ( 5 points)
2. What percentage of scores is between 7.5 and 9.0? ( 5 points)
3. What is the value of \(\mathrm{P}_{20}\), the score separating the bottom \(20 \%\) from the top \(80 \%\) ? (5 points)
4. If 1 scores are randomly selected, what is the probability that their mean is above 9.0 ? (5 points)
5. If 35 scores are randomly selected, what is the probability that their mean is above 9.0? ( 5 points)

Drive-thru service times:

1. The drive-through service times were recorded for 95 randomly selected customers at a Burger King restaurant. Those times had a mean of 127.8 sec and a standard deviation of 60.6 sec. Construct a \(95 \%\) confidence interval estimate of the population mean. ( 10 points)
2. The daily salaries of substitute teachers for 8 local school districts is shown. Find the CI of the mean for the salaries of substitute teachers in the region.( 10 points)
   $60 $56 $6 $55 $70 $55 $60 $55
3. Find sample size necessary to estimate that mean if you want to be confident that the sample mean is within 2 lb of the true population mean. Standard deviation of Population is 12.46 lb. (10 points)
   Solution: The 96% Margin of Error (MOE) is computed as
   \[MOE={{z}_{\alpha /2}}\times \frac{\sigma }{\sqrt{n}}\]
   for \(\alpha =\text{0}\text{.04}\). Therefore, we need the following condition to be satisfied:
   \[MOE={{z}_{\alpha /2}}\times \frac{\sigma }{\sqrt{n}}<\text{2}\,\,\,\,\Leftrightarrow \,\,\,\,\text{2}\text{.05}\times \frac{\text{12}\text{.461}}{\sqrt{n}}<\text{2}\,\,\,\Leftrightarrow \,\,\,n>{{\left( \frac{\text{2}\text{.05}\times \text{12}\text{.461}}{\text{2}} \right)}^{2}}=\text{163}\text{.734}\]
   This means that sample size should be at least n = 164
4. If someone wants to estimate the percentage of households in which a vehicle is owned. How many households must you survey if you want to be \(94 \%\) confident that your sample percentage has a margin of error of three percentage points?

1. Assume that a previous study suggested that vehicles are owned in \(86 \%\) of households.( 5 points)
2. Assume that there is no any previous study.( 5 points)

5. In a study of the Clark method of gender selection, 40 couples tried to have baby girls. Among the 40 babies, \(62.5 \%\) were girls. Construct a \(95 \%\) CI for the proportion of girls from all couples who try to have baby girls with the Clark method of gender selection. (10 points)

Dozenol:

The nighttime cold medicine Dozenol bears a label indicating the presence of \(600 \mathrm{mg}\) of acetaminophen in each fluid ounce of the drug. The Food and Drug Administration randomly selected 40 one-ounce samples and found that the mean acetaminophen content is \(593 \mathrm{mg}\), whereas the population standard deviation is \(21 \mathrm{mg}\). Use \(\alpha=0.01\) and test the claim of the Medassist Pharmaceutical Company that the population mean is different from \(600 \mathrm{mg}\).( Use traditional method and P-value method, 10 points each)

Solution: We are interested in the testing the following null and alternative hypotheses

The population standard deviation \(\sigma \) is known, which means that we are going to use a two-tailed z-test. The z-statistics is computed as

The two-tailed critical z-value for \(\alpha =\text{0}\text{.01}\) is given by

Since \(|z|=\text{2}\text{.108}<\text{2}\text{.58}\), we fail to reject the null hypothesis. The two-tailed p-value is computed as:

Since \(p=\text{0}\text{.035}\) is greater than the significance level \(\alpha =\text{0}\text{.01}\), we fail to reject the null hypothesis. This is equivalent to what we found using the critical value approach. This means that the sample doesn’t provide enough evidence to support the claim that the population mean is different from 600 , at the \[\text{0}\text{.01}\] significance level.

2) Because of the expense involved, car crash tests often use small samples. When 5 BMW cars are crashed under standard conditions, the repair costs (in dollars) are as shown in the accompanying list. Use a 0.05 significance level to test the claim that the mean for all BMW cars is less than $1000.

792 571 904 1147 418

3)Technology is dramatically changing the way we communicate. In 1997, a survey of 880 U.S. households showed that 149 of them use c-mail. Use those sample results to test the claim that less than \(20 \%\) of U.S. households use e-mail. Use a 0.02 significance level. (10 points)

4) Given a data set of 106 healthy body temperatures, where the sample mean was 98.20 and population standard deviation is 0.620, at the 0.05 significance level, test the claim that the mean body temperature of all healthy adults is equal to 98.60. ( Use traditional method and P-value method, 10 points each).

5) One company uses a machine to pour cold medicine into bottles in such a way that the standard deviation of the weight is \(0.15 \mathrm{oz}\). A new machine is tested on 71 bottles, and the standard deviation for this sample is \(0.12 \mathrm{oz}\). The manufactures of new machine, claims that it fills bottles with a lower standard deviation. At the 0.05 significance level, test the claim. (10 points)

4. Insurance Company

An insurance company wants to determine the strength of the relationship between the number of hours a person works per week and the number of injuries or accidents that person has over a period of one week.

The data follow. Show formulas and all your works

Compute the correlation coefficient. (2.5 points)

Find critical value when \(\alpha=0.05\), is there significant linear correlation between \(\mathrm{x}\) and \(\mathrm{y}\) ? (2.5 points)

Determine the regression equation(2.5 points)

Using the regression equation, what is the best predicted value of the number of accidents for a person who work 10 hours per week? (2.5 points)

Price: $39.85

Solution: The downloadable solution consists of 18 pages, 2185 words.
Deliverable: Word Document