Standard Normal Distribution: Videos & Practice Problems

In the study of discrete random variables and their probability distributions, understanding variance and standard deviation is crucial as they indicate how spread out the outcomes of an experiment are. The process for calculating these measures involves specific equations tailored for probability distributions, which may initially seem complex but can be simplified through systematic steps.

To find the variance (\( \sigma^2 \)) of a discrete random variable, you can use the formula:

$$ \sigma^2 = \sum (x^2 \cdot P(x)) - \mu^2 $$

Here, \( x \) represents the possible outcomes, \( P(x) \) is the probability of each outcome, and \( \mu \) is the mean (or expected value) of the distribution. The mean can be calculated as:

$$ \mu = \sum (x \cdot P(x)) $$

To organize the calculations, it is helpful to create a table with columns for \( x \), \( P(x) \), \( x \cdot P(x) \), and \( x^2 \cdot P(x) \). This structured approach allows for easier computation of the necessary sums.

For example, if you have a probability distribution for the number of children in a household, you would first calculate the mean, which might be given as 1.1. Squaring this value gives \( \mu^2 = 1.21 \). Next, you would compute the sum of \( x^2 \cdot P(x) \) for each outcome, which might total 1.60. The variance is then found by subtracting the mean squared from this sum:

$$ \sigma^2 = 1.60 - 1.21 = 0.39 $$

Once the variance is determined, the standard deviation (\( \sigma \)) can be easily calculated as the square root of the variance:

$$ \sigma = \sqrt{\sigma^2} $$

In this case, taking the square root of 0.39 yields approximately 0.62, which can be rounded to 0.6 when considering the context of the data (e.g., kids per household).

Ultimately, the variance and standard deviation provide valuable insights into the variability of the data, with a mean of 1.1 and a standard deviation of about 0.6 indicating the expected spread of outcomes in this scenario. Understanding these concepts is essential for analyzing and interpreting data in probability and statistics.

In statistics, calculating probabilities using z-scores is a fundamental skill, particularly when working with the standard normal distribution. A common scenario involves finding the probability that a value falls between two z-scores. This process requires understanding how to interpret z-tables and perform basic arithmetic operations with the probabilities obtained from them.

To find the probability between two z-scores, such as -0.64 and 2.27, the first step is to identify the cumulative probabilities associated with each z-score using a z-table. The cumulative probability for a z-score represents the area under the curve to the left of that z-score. For instance, if the cumulative probability for z = -0.64 is 0.2611, this indicates that approximately 26.11% of the data falls below this z-score. Conversely, for z = 2.27, if the cumulative probability is 0.9884, it signifies that about 98.84% of the data is below this z-score.

To find the probability of a value falling between these two z-scores, the procedure involves subtracting the cumulative probability of the lower z-score from that of the higher z-score. This can be expressed mathematically as:

$$ P(-0.64 < Z < 2.27) = P(Z < 2.27) - P(Z < -0.64) $$

Substituting the values from the z-table, we have:

$$ P(-0.64 < Z < 2.27) = 0.9884 - 0.2611 = 0.7273 $$

This result indicates that there is a 72.73% probability that a randomly selected value from the standard normal distribution will fall between the z-scores of -0.64 and 2.27. Understanding this process is crucial for interpreting data and making informed decisions based on statistical analysis.

In statistics, understanding how to find z-scores from given probabilities under the standard normal curve is essential. This process often involves using a z-table, which provides cumulative probabilities associated with z-scores. When given a probability, the first step is to sketch the normal distribution curve to visualize the area represented by that probability. This helps in determining whether the z-score will be positive or negative based on the area provided.

For instance, if you are given a cumulative probability of 0.8051, this indicates that the area to the left of the z-score is 0.8051. Since this value is greater than 0.5, you can expect the z-score to be a positive number. To find the corresponding z-score, you look up the value 0.8051 directly in the z-table. The z-table lists cumulative probabilities, and by locating 0.8051, you can determine that the associated z-score is approximately 0.86.

However, when the problem specifies an area to the right of a z-score, such as a probability of 0.67, the approach changes slightly. In this case, you first need to find the area to the left of the z-score by subtracting the given probability from 1. Thus, you would calculate 1 - 0.67, resulting in an area of 0.33. This area is what you will look up in the z-table. Upon finding 0.33 in the table, you can determine that the corresponding z-score is approximately -0.44, indicating a small negative value, which aligns with the expectation based on the area provided.

In summary, when working with areas to the left of a z-score, you can directly use the given probability in the z-table. For areas to the right, always convert the probability to the left by subtracting from 1 before looking it up in the z-table. This systematic approach ensures accurate identification of z-scores based on the probabilities provided.