Table of contents
- 1. Intro to Stats and Collecting Data55m
- 2. Describing Data with Tables and Graphs1h 55m
- 3. Describing Data Numerically1h 45m
- 4. Probability2h 16m
- 5. Binomial Distribution & Discrete Random Variables2h 33m
- 6. Normal Distribution and Continuous Random Variables1h 38m
- 7. Sampling Distributions & Confidence Intervals: Mean1h 3m
- 8. Sampling Distributions & Confidence Intervals: Proportion1h 12m
- 9. Hypothesis Testing for One Sample1h 1m
- 10. Hypothesis Testing for Two Samples2h 8m
- 11. Correlation48m
- 12. Regression1h 4m
- 13. Chi-Square Tests & Goodness of Fit1h 20m
- 14. ANOVA1h 0m
6. Normal Distribution and Continuous Random Variables
Standard Normal Distribution
Problem 6.R.6b
Textbook Question
Mensa Membership in Mensa requires a score in the top 2% on a standard intelligence test. The Wechsler IQ test is designed for a mean of 100 and a standard deviation of 15, and scores are normally distributed.
b. If 4 randomly selected adults take the Wechsler IQ test, find the probability that their mean score is at least 131.

1
Step 1: Understand the problem. The question asks for the probability that the mean IQ score of 4 randomly selected adults is at least 131. Since the scores are normally distributed, we can use properties of the normal distribution to solve this.
Step 2: Calculate the standard error of the mean (SEM). The SEM is given by the formula: , where is the population standard deviation (15 in this case) and is the sample size (4 in this case).
Step 3: Standardize the mean score of 131 using the z-score formula: , where is the sample mean (131), is the population mean (100), and is the standard error of the mean calculated in Step 2.
Step 4: Use the z-score obtained in Step 3 to find the corresponding probability from the standard normal distribution table. This will give the probability that the mean score is less than 131.
Step 5: Subtract the probability obtained in Step 4 from 1 to find the probability that the mean score is at least 131. This is because the question asks for the probability of the mean score being at least 131, which corresponds to the upper tail of the 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 the context of the Wechsler IQ test, scores are normally distributed with a mean of 100 and a standard deviation of 15, which allows us to use the properties of the normal distribution to calculate probabilities related to IQ scores.
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Central Limit Theorem
The Central Limit Theorem states that the sampling distribution of the sample mean will be normally distributed, regardless of the shape of the population distribution, provided the sample size is sufficiently large (typically n ≥ 30). In this case, since we are dealing with a sample of 4 adults, we can still apply the theorem to approximate the distribution of the sample mean, allowing us to calculate the probability of their mean score being at least 131.
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Calculating the Mean
Z-Score
A Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, expressed in terms of standard deviations from the mean. To find the probability that the mean score of the 4 adults is at least 131, we first calculate the Z-score for 131 using the formula Z = (X - μ) / (σ/√n), where X is the score of interest, μ is the mean, σ is the standard deviation, and n is the sample size.
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