3. Describing Data Numerically

Standard Deviation

3. Describing Data Numerically

Standard Deviation

2:53 minutes

Problem 3.3.4

Textbook Question

z Scores If your score on your next statistics test is converted to a z score, which of these z scores would you prefer: -2.00, -1.00, 0, 1.00, 2.00? Why?

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Understand the concept of a z-score: A z-score measures how many standard deviations a data point is from the mean. A positive z-score indicates a value above the mean, while a negative z-score indicates a value below the mean.

Interpret the given z-scores: A z-score of -2.00 means the score is 2 standard deviations below the mean, -1.00 is 1 standard deviation below the mean, 0 is exactly at the mean, 1.00 is 1 standard deviation above the mean, and 2.00 is 2 standard deviations above the mean.

Consider the context of the problem: Since the question is about a test score, a higher z-score is preferable because it indicates a better performance relative to the average (mean) score.

Compare the z-scores: Among the given z-scores (-2.00, -1.00, 0, 1.00, 2.00), the z-score of 2.00 is the highest, meaning it represents the best performance relative to the mean.

Conclude: You would prefer a z-score of 2.00 because it indicates that your test score is 2 standard deviations above the mean, which is the best performance among the options provided.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Z Score

A z score, or standard score, indicates how many standard deviations a data point is from the mean of a dataset. It is calculated by subtracting the mean from the data point and then dividing by the standard deviation. Z scores allow for comparison between different datasets by standardizing scores, making it easier to understand relative performance.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range. Understanding standard deviation is crucial for interpreting z scores, as it affects how far a score is from the mean.

Normal Distribution

Normal distribution is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In a normal distribution, z scores can be used to determine the probability of a score occurring within a certain range. This concept is essential for understanding the implications of different z scores in terms of performance relative to peers.

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Textbook Question

Estimating Standard Deviation with the Range Rule of Thumb. In Exercises 29–32, refer to the data in the indicated exercise. After finding the range of the data, use the range rule of thumb to estimate the value of the standard deviation. Compare the result to the standard deviation computed using all of the data.

Body Temperatures Refer to Data Set 5 “Body Temperatures” in Appendix B and use the body temperatures for 12:00 AM on day 2.

Textbook Question

Finding Standard Deviation from a Frequency Distribution. In Exercises 37–40, refer to the frequency distribution in the given exercise and compute the standard deviation by using the formula below, where x represents the class midpoint, f represents the class frequency, and n represents the total number of sample values. Also, compare the computed standard deviations to these standard deviations obtained by using Formula 3-4 with the original list of data values: (Exercise 37) 18.5 minutes; (Exercise 38) 36.7 minutes; (Exercise 39) 6.9 years; (Exercise 40) 20.4 seconds.

Standard deviation for frequency distribution

Textbook Question

The Empirical Rule Based on Data Set 1 “Body Data” in Appendix B, blood platelet counts of women have a bell-shaped distribution with a mean of 255.1 and a standard deviation of 65.4. (All units are 1000 cells/) Using the empirical rule, what is the approximate percentage of women with platelet counts

a. within 2 standard deviations of the mean, or between 124.3 and 385.9?

Textbook Question

Why Divide by ? Let a population consist of the values 9 cigarettes, 10 cigarettes, and 20 cigarettes smoked in a day (based on data from the California Health Interview Survey). Assume that samples of two values are randomly selected with replacement from this population. (That is, a selected value is replaced before the second selection is made.)

a. Find the variance of the population {9 cigarettes, 10 cigarettes, 20 cigarettes}.

Textbook Question

Standard deviation for frequency distribution

Textbook Question

b. After listing the nine different possible samples of two values selected with replacement, find the sample variance (which includes division by ) for each of them; then find the mean of the nine sample variances s2.

Textbook Question

c. For each of the nine different possible samples of two values selected with replacement, find the variance by treating each sample as if it is a population (using the formula for population variance, which includes division by n); then find the mean of those nine population variances.

Textbook Question

d. Which approach results in values that are better estimates of part (b) or part (c)? Why? When computing variances of samples, should you use division by n or

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