Table of contents

1. Intro to Stats and Collecting Data1h 14m
2. Describing Data with Tables and Graphs1h 55m
3. Describing Data Numerically2h 5m
4. Probability2h 16m
5. Binomial Distribution & Discrete Random Variables3h 6m
6. Normal Distribution and Continuous Random Variables2h 11m
7. Sampling Distributions & Confidence Intervals: Mean3h 23m
8. Sampling Distributions & Confidence Intervals: Proportion1h 12m
- Sampling Distribution of Sample Proportion
  29m
- Confidence Intervals for Population Proportion
  42m
9. Hypothesis Testing for One Sample3h 29m
10. Hypothesis Testing for Two Samples4h 50m
11. Correlation1h 6m
12. Regression1h 50m
13. Chi-Square Tests & Goodness of Fit1h 57m
14. ANOVA1h 57m

6. Normal Distribution and Continuous Random Variables

Standard Normal Distribution

Statistics

6. Normal Distribution and Continuous Random Variables

Standard Normal Distribution

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1:19 minutes

Problem 6.2.2b

Textbook Question

Hershey Kisses Based on Data Set 38 “Candies” in Appendix B, weights of the chocolate in Hershey Kisses are normally distributed with a mean of 4.5338 g and a standard deviation of 0.1039 g

b. What is the value of the median?

Verified step by step guidance

Step 1: Understand the relationship between the mean and the median in a normal distribution. In a normal distribution, the mean, median, and mode are all equal because the distribution is symmetric.

Step 2: Identify the given parameters of the normal distribution. The problem states that the mean weight of Hershey Kisses is 4.5338 g and the standard deviation is 0.1039 g.

Step 3: Recall that the median in a normal distribution is equal to the mean. This is a property of the normal distribution due to its symmetry.

Step 4: Conclude that the median weight of Hershey Kisses is the same as the mean weight, which is 4.5338 g.

Step 5: No further calculations are needed because the median is directly determined by the mean in a normal distribution.

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Key Concepts

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

Normal Distribution

Normal distribution is a probability distribution that is symmetric about the mean, indicating that data near the mean are more frequent in occurrence than data far from the mean. In a normal distribution, the mean, median, and mode are all equal, which simplifies the calculation of the median when the distribution is known.

Mean

The mean is the average of a set of values, calculated by summing all the values and dividing by the number of values. In the context of a normal distribution, the mean serves as the central point around which the data is distributed, and it is crucial for understanding the overall distribution of the dataset.

Median

The median is the middle value of a dataset when it is ordered from least to greatest. For a normal distribution, the median is equal to the mean, making it straightforward to determine in this case. It represents the point at which half of the data points fall below and half fall above.

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Textbook Question

Critical Values. In Exercises 41–44, find the indicated critical value. Round results to two decimal places.

z0.90

Textbook Question

Basis for the Range Rule of Thumb and the Empirical Rule. In Exercises 45–48, find the indicated area under the curve of the standard normal distribution; then convert it to a percentage and fill in the blank. The results form the basis for the range rule of thumb and the empirical rule introduced in Section 3-2.

About __ % of the area is between z = -3.5 and z = 3.5 (or within 3.5 standard deviation of the mean).

Textbook Question

Significance For bone density scores that are normally distributed with a mean of 0 and a standard deviation of 1, find the percentage of scores that are

c. not significant (or less than 2 standard deviations away from the mean).

Textbook Question

Hershey Kisses Based on Data Set 38 “Candies” in Appendix B, weights of the chocolate in Hershey Kisses are normally distributed with a mean of 4.5338 g and a standard deviation of 0.1039 g.

a. What are the values of the mean and standard deviation after converting all weights of Hershey Kisses to z scores using z = (x - μ)/σ ?

b. The original weights are in grams. What are the units of the corresponding z scores?

Textbook Question

Hershey Kisses Based on Data Set 38 “Candies” in Appendix B, weights of the chocolate in Hershey Kisses are normally distributed with a mean of 4.5338 g and a standard deviation of 0.1039 g

d. What is the value of the variance?

Textbook Question

Eye Color Based on a study by Dr. P. Sorita at Indiana University, assume that 12% of us have green eyes. In a study of 650 people, it is found that 86 of them have green eyes.

b. Is 86 people with green eyes significantly high?

Textbook Question

Mendelian Genetics When Mendel conducted his famous genetics experiments with peas, one sample of offspring consisted of 929 peas, with 705 of them having red flowers. If we assume, as Mendel did, that under these circumstances, there is a 3/4 probability that a pea will have a red flower, we would expect that 696.75 (or about 697) of the peas would have red flowers, so the result of 705 peas with red flowers is more than expected.

a. If Mendel’s assumed probability is correct, find the probability of getting 705 or more peas with red flowers.

Textbook Question

Bone Density Test A bone mineral density test is used to identify a bone disease. The result of a bone density test is commonly measured as a z score, and the population of z scores is normally distributed with a mean of 0 and a standard deviation of 1.

e. If the mean bone density test score is found for 9 randomly selected subjects, find the probability that the mean is greater than 0.23.

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