High School

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**For each of the following, identify if it is descriptive or inferential:**

a) The class average on last Friday's exam was 71%.
b) The class average on all exams in my course based on the last 2 years of data is 71%.
c) The percent of people that plan to vote for Candidate Ashkenian in the next election.
d) The percent of people that did vote for Candidate Ashkenian on election day.

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**Chapter 3**

1. A survey was conducted asking how many times a month they eat at restaurants. The following results were obtained:

1a) Calculate the mean
1b) Calculate the median
1c) Calculate the mode
1d) Calculate the third quartile

2. Given the following sample: 32, 76, 44, 58, 32, 19, 63, 51

2a) Compute the range
2b) Compute the interquartile range
2c) Compute the variance
2d) Compute the standard deviation

3. Given the following values: 3, 6, 7, 9, 10

3a) Calculate the mean
3b) Calculate the standard deviation
3c) Calculate the z-score for each value

4. I did a study and found that my sample variable has a mean of 100 and a standard deviation of 10. Use Chebyshev's Theorem to determine the percent of data in each range:

4a) 80−120
4b) 75-125
4c) 50-150

I further analyzed my dataset and discovered that it is normally distributed.

4d) What percent of my values would be between 85-115 in a normal distribution?

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**Chapter 6**

1. Using the Normal Distribution (z), calculate the following probabilities:

1a) P(0.35)

Answer :

(a) Descriptive,(b) Inferential,(c) Inferential,(d) Descriptive and 2. Range = 57, Interquartile Range (IQR) = 28.5, Variance = 392.1875, Standard Deviation = 19.8, Mean = 45.5 and 3. Variance = 4.8, Standard Deviation = 2.19 , Mean = 7

Z-scores = Calculated by subtracting the mean from each value and dividing by the standard deviation. When the data is normally distributed, approximately 68% of values fall within one standard deviation of the mean. Therefore, the percent of values between 85-115 would be approximately 68%.

a) Descriptive - the class average on last Friday's exam was 71%.

b) Inferential - the class average on all exams in my course based on the last 2 years of data is 71%.

c) Inferential - the percent of people that plan to vote for Candidate Ashkenian in the next election.

d) Descriptive - the percent of people that did vote for Candidate Ashkenian on election day.

2.Given the sample: 32, 76, 44, 58, 32, 19, 63, 51.

a) Range = Maximum value - Minimum value = 76 - 19 = 57.

b) To compute the interquartile range (IQR), we first need to find the first quartile (Q1) and the third quartile (Q3) by arranging the data in ascending order: 19, 32, 32, 44, 51, 58, 63, 76.

Q1 = (32 + 32) / 2 = 32.

Q3 = (58 + 63) / 2 = 60.5.

IQR = Q3 - Q1 = 60.5 - 32 = 28.5.

c) To compute the variance, we first calculate the mean (average) of the data: (32 + 76 + 44 + 58 + 32 + 19 + 63 + 51) / 8 = 45.5.

Next, we subtract the mean from each data point, square the result, and calculate the average of the squared differences:

Variance = ((32 - 45.5)^2 + (76 - 45.5)^2 + (44 - 45.5)^2 + (58 - 45.5)^2 + (32 - 45.5)^2 + (19 - 45.5)^2 + (63 - 45.5)^2 + (51 - 45.5)^2) / 8 = 392.1875.

d) The standard deviation is the square root of the variance: √392.1875 ≈ 19.8.

3.Given the values: 3, 6, 7, 9, 10.

a) Mean = (3 + 6 + 7 + 9 + 10) / 5 = 7.

b) To calculate the standard deviation, we first find the variance:

Variance = [(3 - 7)^2 + (6 - 7)^2 + (7 - 7)^2 + (9 - 7)^2 + (10 - 7)^2] / 5 = 4.8.

Standard Deviation = √4.8 ≈ 2.19.

c) To calculate the z-score for each value, we subtract the mean from each value and divide by the standard deviation:

For 3: (3 - 7) / 2.19 ≈ -1.83

For 6: (6 - 7) / 2.19 ≈ -0.46

For 7: (7 - 7) / 2.19 ≈ 0

For 9: (9 - 7) / 2.19 ≈ 0.92

For 10: (10 - 7) / 2.19 ≈ 1.37

4. Given that the sample variable has a mean of 100 and a standard deviation of 10, we can use Chebyshev's Theorem and the characteristics of the normal distribution to determine the percent of data within specific ranges.

(a) Range: 80-120.

Chebyshev's Theorem guarantees that at least (1 - 1/k^2) of the data falls within k standard deviations from the mean. In this case, k = (120 - 100) / 10 = 2.

Thus, at least 1 - 1/2^2 = 1 - 1/4 = 75% of the data falls within the range 80-120.

(b) Range: 75-125.

Using the same reasoning as above, k = (125 - 100) / 10 = 2.5.

Therefore, at least 1 - 1/2.5^2 = 1 - 1/6.25 = 84% of the data falls within the range 75-125.

(c) Range: 50-150.

Here, k = (150 - 100) / 10 = 5.

Hence, at least 1 - 1/5^2 = 1 - 1/25 = 96% of the data falls within the range 50-150.

(d) Since the variable is normally distributed, we can use the properties of the normal distribution. In a normal distribution, approximately 68% of the data falls within one standard deviation from the mean. Therefore, the percentage of values between 85-115 would also be approximately 68%.

Learn more about normal distribution here:

https://brainly.com/question/14916937

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