This workbook accompanies the Normal Distribution The Normal Distribution lesson and is built for practice with a calculator and a standard normal table. The ten exercises follow the lesson’s order, starting with random variables and the population mean and variance, then moving through standardization, reading probabilities from the z-table, working backwards from a probability to a value, and finally the sampling distribution of the mean and the central limit theorem. Each exercise isolates one skill so you can see precisely where your method needs work.
Attempt every exercise before reading the solutions, writing out each step rather than skipping to the answer. The solutions name the formula, show the arithmetic, and interpret the result, so the goal is to check your method against the standard one. The standard normal values you need are supplied inside each exercise, so you can focus on the technique rather than table lookups.
Part One: The Exercises
Exercise 1 (Discrete or continuous). Classify each of the following random variables as discrete or continuous: the number of emails received in a day, the exact weight of a parcel, the number of defective items in a batch, and the time taken to run a race. Explain the rule you used.
Exercise 2 (Expected value). A discrete random variable X takes the values 0, 1, 2, and 3 with probabilities 0.1, 0.3, 0.4, and 0.2 respectively. Compute the population mean E(X).
Exercise 3 (Variance and standard deviation). Using the same distribution as Exercise 2, compute the variance and standard deviation of X with the shortcut formula.
Exercise 4 (Standardization). A variable X is normally distributed with mean 50 and variance 100. Compute the z-score for an observation of X = 65 and for an observation of X = 42.
Exercise 5 (Cumulative and upper-tail probability). For the standard normal distribution, given that the cumulative probability up to 1.5 is 0.9332, find the probability that Z is at most 1.5 and the probability that Z exceeds 1.5.
Exercise 6 (Symmetry of the standard normal). Given that the cumulative probability up to 0.8 is 0.7881, use the symmetry of the standard normal distribution to find the probability that Z is less than -0.8.
Exercise 7 (Interval probability for a normal variable). A variable X is normally distributed with mean 100 and variance 225. Find the probability that X lies between 85 and 115, using the value that the cumulative probability up to 1.0 is 0.8413.
Exercise 8 (Working backwards from a probability). Exam scores are normally distributed with mean 500 and standard deviation 100. Find the score below which 95% of candidates fall, using the critical value 1.6449 for the 95th percentile.
Exercise 9 (The empirical rule). Adult heights in a population are normally distributed with mean 170 cm and standard deviation 5 cm. Using the properties of the normal distribution, find the range that contains the middle 95% of heights, and find the probability that a randomly chosen adult is taller than 175 cm, given that the cumulative probability up to 1.0 is 0.8413.
Exercise 10 (Sampling distribution and the CLT). A population has mean 100 and standard deviation 20. A random sample of 25 observations is taken. State the sampling distribution of the sample mean, then find the probability that the sample mean is less than 95, using the value that the cumulative probability up to 1.25 is 0.8944.
Part Two: Worked Solutions
Solution 1. A discrete variable takes countable, separate values, while a continuous variable can take any value within a range. The number of emails and the number of defective items are discrete, because they are counts that jump from one whole number to the next. The exact weight of a parcel and the time taken to run a race are continuous, because they can take any value on a scale, limited only by the precision of measurement.
Solution 2. The population mean is the probability-weighted average of all possible values.
Adding the terms gives 0 plus 0.3 plus 0.8 plus 0.6.
The expected value of X is 1.7.
Solution 3. The shortcut formula subtracts the square of the mean from the expected value of the square. First find the expected value of X squared.
Then apply the shortcut.
The standard deviation is the square root of the variance.
The variance is 0.81 and the standard deviation is 0.9.
Solution 4. The variance is 100, so the standard deviation is 10, and standardization subtracts the mean and divides by the standard deviation.
For the second observation the same formula applies.
The z-scores are 1.5 and -0.8.
Solution 5. The cumulative probability up to a value is exactly the probability of being at or below it.
The upper tail is the complement, since the total area under the curve is 1.
The probabilities are 0.9332 and 0.0668.
Solution 6. By symmetry, the area below a negative value equals the area above the matching positive value.
The probability that Z is less than -0.8 is 0.2119.
Solution 7. The variance is 225, so the standard deviation is 15, and an interval probability is the difference of two cumulative probabilities after standardizing both endpoints.
Using symmetry, the lower term is 1 minus 0.8413, which is 0.1587.
The probability is 0.6826, which is the familiar result that about 68% of a normal distribution lies within one standard deviation of the mean.
Solution 8. Working backwards starts from the z-score of the desired percentile and converts it to an X value by reversing standardization. The 95th percentile corresponds to a z-score of 1.6449.
About 95% of candidates score below 664.5.
Solution 9. By the properties of the normal distribution, the middle 95% lies within roughly two standard deviations of the mean.
So the middle 95% of heights fall between 160 cm and 180 cm. For the probability of exceeding 175 cm, standardize first, which gives a z-score of 1.0, then take the upper tail.
About 15.9% of adults are taller than 175 cm.
Solution 10. By the central limit theorem, the sample mean is normally distributed with the same mean as the population and a variance reduced by the sample size.
The standard error is the standard deviation divided by the square root of the sample size.
Standardize the sample mean of 95 using this standard error.
Then read off the lower-tail probability using symmetry.
The probability that the sample mean is below 95 is about 0.1056.
How to Get the Most From This Workbook
The thread running through these exercises is standardization. Almost every probability question about a normal variable becomes easy once you convert it to a z-score, because the z-table then does the rest. Notice that the only thing that changes between a single observation and a sample mean is which standard deviation goes in the denominator: the population standard deviation for one observation, and the standard error for a sample mean. Working backwards from a probability simply runs the same conversion in reverse. Practise that one move until it is automatic, and the normal distribution stops being a table-lookup chore and becomes a single reliable technique applied again and again.
See you soon.
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