Gifted/ Accelerated Math 3

Gifted/ Accelerated Math 3 Name:________________________

Statistics Quiz 1 Review Date:_________________________

1. The mean lifetime of all home heaters of a certain brand is 9.5 years. A random sample of 30 home heaters is taken and the average lifetime of these heaters is 8.2 years. What is the population?

a. All home heaters that have failed.

b. All home heaters with a lifetime of 9.5 years.

c. All home heaters that have been sold.

d. All home heaters ever produced.

2. Based on a random sample of 6 households, the number of gallons of water consumed per day is listed. Obtain the sample mean and sample standard deviation.

10 6 8 6 5 12

3. At a company, the mean starting salary is $31,000 and the standard deviation is $1000. What’s the z-score for someone whose starting salary is $30,000?

4. The value of land in a certain area appears to be bell-shaped with a mean of $1000 per acre and standard deviation $200. Estimate the percentage of farms whose land value is between $800 and $1200.

5. The distribution of the heights of students in a large class is roughly bell-shaped. Moreover, the average height is 68 inches, and approximately 95% of the heights are between 62 and 74 inches. Thus, the standard deviation of the height distribution is approximately equal to:

a. 2

b. 3

c. 6

d. 9

e. 12

6. The grade point averages of students at the University of Manitoba are approximately normally distributed with mean equal to 2.4 and standard deviation equal to 0.8. What fraction of the students will possess a grade point average in excess of 3.0 ?

7. Grades on a Chemistry test follow a normal distribution with a mean of 65 and a standard deviation of 12. Approximately what percentage of the students have scores below 50?

8. Sketch a normal distribution and plot and label the properties of the Empirical Rule.

9. One of the side effects of flooding a lake in northern boreal forest areas (e.g. for a hydro-electric project) is that mercury is leached from the soil, enters the food chain, and eventually contaminates the fish. The concentration in fish will vary among individual fish because of differences in eating patterns, movements around the lake, etc. Suppose that the concentration of mercury in individual fish follows an approximate normal distribution with a mean of 0.25 ppm and a standard deviation of 0.08 ppm. Fish are safe to eat if the mercury level is below 0.30 ppm. What proportion of fish are safe to eat?

10. How does the standard deviation of a sample proportion change as the sample size grows larger?

11. If we wanted to cut the standard deviation of a sample proportion by a fifth what would we need to do in terms of our sample size?

12. About 65% of all university students at a large university (approx. 10,000 students) belong to the student loan scheme. Consider a random sample of 50 students. Let be the proportion of these 50 students who belong to the student loan scheme.

a. In words, describe p.

b. State the distribution (mean and standard deviation) of .

c. Explain why you can use the formula for standard deviation of in this situation.

d. What is the probability that the sample proportion is more than 70%?

e. What is the probability that the sample proportion is between 0.45 and 0.55?

13. A poll conducted by students in the Introduction to Statistics course surveyed the freshmen class (approximately 5000 students) on how many freshmen took the SAT or a different test as their college entrance exam. Their study found that 83% took the SAT and 17% took a different test.

A curious history teacher decided to poll his 100-person freshmen class to compare his results of his class to the results provided from the study. According to the Central Limit Theorem was his sample size large enough for his results to be valid?

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Central Limit Theorem:

If a random sample of n observation is selected from any population, then, when the sample size is sufficiently large (n>=30) the sampling distribution of the mean tends to approximate the normal distribution. The larger the sample size, n, the better will be the normal approximation to the sampling distribution of the mean. Then, again in this case it can be shown that the mean of the sample means is same as population mean.

The real advantage of the central limit theorem is that sample data drawn from populations not normally distributed or from populations of unknown shape also can be analyzed by using the normal distribution, because the sample means are normally distributed for sample sizes of n>=30.

Column 1 of the following figure shows four different population distributions. Each ensuing column displays the shape of the distribution of the sample means for a particular sample size. Note that the distribution of the sample means begins to approximate the normal curve as the sample size, n, gets larger.

Since the central limit theorem states that sample means are normally distributed regardless of the shape of the population for large samples and for any sample size with normally distributed population, thus sample means can be analyzed by using Z scores.

Gifted/ Accelerated Math 3 Name:________________________

Statistics Quiz 1 Review - Solutions Date:_________________________

1. The mean lifetime of all home heaters of a certain brand is 9.5 years. A random sample of 30 home heaters is taken and the average lifetime of these heaters is 8.2 years. What is the population?

a. All home heaters that have failed.

b. All home heaters with a lifetime of 9.5 years.

c. All home heaters that have been sold.

d. All home heaters ever produced.

2. Based on a random sample of 6 households, the number of gallons of water consumed per day is listed. Obtain the sample mean and sample standard deviation.

10 6 8 6 5 12

3. At a company, the mean starting salary is $31,000 and the standard deviation is $1000. What’s the z-score for someone whose starting salary is $30,000?

$800 & $1200 are both 1 standard deviation away from the mean, therefore the percentage of farms who land is between them is 68% by the Empirical Rule

e. 2

f. 3

g. 6

h. 9

i. 12

, next find the probability that has the z-score of 0.75 from the z-score chart. The area below z-score 0.75 is 0.7734, but we are looking for the people who have a GPA greater than 3.0. We take 1 – 0.7734 = 0.2266.

Therefore, 22.66% of the people at the university will have a GPA greater than 3.0.

7. Grades on a Chemistry test follow a normal distribution with a mean of 65 and a standard deviation of 12. Approximately what percentage of the students have scores below 50?

, next find the probability that has a z-score of -1.25 from the z-score chart. The area below (or to the left) is 0.1056. Therefore, 10.56% of the grades were below 50.

8. Sketch a normal distribution and plot and label the properties of the Empirical Rule.

, next find the probability that falls below 0.30 ppm by looking up 0.625 in the z-score chart. The probability is either 0.7324 or 0.7357 (depending if you used 0.62 or 0.63). Therefore you 73.24% or 73.57% of the fish are safe to eat.

10. How does the standard deviation of a sample proportion change as the sample size grows larger?

As sample size grows larger the standard deviation will decrease.

11. If we wanted to cut the standard deviation of a sample proportion by a fifth what would we need to do in terms of our sample size?

Your sample size would need to be 25 times as large to cut the standard deviation by a fifth.

j. In words, describe p.

The proportion of university students who belong to the student loan scheme.

k. State the distribution (mean and standard deviation) of .

l. Explain why you can use the formula for standard deviation of in this situation.

The population is at least ten times the sample size.

m. What is the probability that the sample proportion is more than 70%?

, next find the probability that is less than 70% using the z-score chart, 0. 7704. Since we want to know the proportion that is more than 70% we take 1 – 0.7704 = 0.2296. Therefore, the proportion of students more than 70% is 22.96%.

n. What is the probability that the sample proportion is between 0.45 and 0.55?

To find the area between 0.45 and 0.55 find the probabilities associated with both z-score. We have 0.0694 – 0.0015 = 0.0679. Therefore, 6.79% falls between 0.45 & 0.55.

100(0.83) = 83 & 100(1 – 0.83) = 17 à therefore the sample size is large enough to use the Central Limit Theorem.