Question

For a normal distribution where the mean is 50 and the standard deviation is 10 , what percentage of the area is a. between the scores of 40 and \(47 ?\) b. above a score of 47 ? \(\mathbf{c}\). below a score of 53 ? d. between the scores of 35 and 65 ? \(e_{*}\) above a score of 72 ? \(\mathbf{f}\). below a score of 31 and above a score of 69 ? g. between the scores of 55 and 62 ? h. between the scores of 32 and 47 ?

Short answer

a. 22.34%b. 61.79%c. 61.79%d. 86.64%e. 1.39%f. 5.74%g. 19.34%h. 34.62%

Step-by-step solution

- Convert raw scores to z-scores

For each raw score, convert it to a z-score using the formula: \[ z = \frac{x - \text{mean}}{\text{standard deviation}} \] Here, the mean is 50 and the standard deviation is 10.

- Calculate z-scores for each part

a. Scores 40 and 47: \[ z_{40} = \frac{40 - 50}{10} = -1 \] \[ z_{47} = \frac{47 - 50}{10} = -0.3 \] b. Score 47: \[ z_{47} = -0.3 \] c. Score 53: \[ z_{53} = \frac{53 - 50}{10} = 0.3 \] d. Scores 35 and 65: \[ z_{35} = \frac{35 - 50}{10} = -1.5 \] \[ z_{65} = \frac{65 - 50}{10} = 1.5 \] e. Score 72: \[ z_{72} = \frac{72 - 50}{10} = 2.2 \] f. Scores 31 and 69: \[ z_{31} = \frac{31 - 50}{10} = -1.9 \] \[ z_{69} = \frac{69 - 50}{10} = 1.9 \] g. Scores 55 and 62: \[ z_{55} = \frac{55 - 50}{10} = 0.5 \] \[ z_{62} = \frac{62 - 50}{10} = 1.2 \] h. Scores 32 and 47: \[ z_{32} = \frac{32 - 50}{10} = -1.8 \] \[ z_{47} = -0.3 \]

- Use standard normal distribution table

Look up these z-scores in the standard normal distribution table to find the areas under the curve. These areas represent the probabilities. Recall: - Area to the left of -1.0: 0.1587- Area to the left of -0.3: 0.3821- Area to the left of 0.3: 0.6179- Area to the left of -1.5: 0.0668- Area to the left of 1.5: 0.9332- Area to the left of 2.2: 0.9861- Area to the left of -1.9: 0.0287- Area to the left of 1.9: 0.9713- Area to the left of 0.5: 0.6915- Area to the left of 1.2: 0.8849- Area to the left of -1.8: 0.0359

- Calculate percentage areas

a. Between scores 40 and 47: \( P(40 < x < 47) = P(z < -0.3) - P(z < -1.0) = 0.3821 - 0.1587 = 0.2234 \) b. Above score 47: \( P(x > 47) = 1 - P(z < -0.3) = 1 - 0.3821 = 0.6179 \) c. Below score 53: \( P(x < 53) = P(z < 0.3) = 0.6179 \) d. Between scores 35 and 65: \( P(35 < x < 65) = P(z < 1.5) - P(z < -1.5) = 0.9332 - 0.0668 = 0.8664 \) e. Above score 72: \( P(x > 72) = 1 - P(z < 2.2) = 1 - 0.9861 = 0.0139 \) f. Below score 31 and above score 69: \( P(x < 31) + P(x > 69) = 0.0287 + (1 - 0.9713) = 0.0287 + 0.0287 = 0.0574 \) g. Between scores 55 and 62: \( P(55 < x < 62) = P(z < 1.2) - P(z < 0.5) = 0.8849 - 0.6915 = 0.1934 \) h. Between scores 32 and 47: \( P(32 < x < 47) = P(z < -0.3) - P(z < -1.8) = 0.3821 - 0.0359 = 0.3462 \)

Key concepts

Understanding z-scores

Z-scores are a way to describe how far a specific score is from the mean of a distribution, in terms of the number of standard deviations. Z-scores help us understand the position of a value within a normal distribution. To calculate a z-score, use the formula: \( z = \frac{x - \text{mean}}{\text{standard deviation}} \). In this specific exercise, the mean (average) is 50 and the standard deviation is 10. For example, to find the z-score for 40, you compute: \( z_{40} = \frac{40 - 50}{10} = -1 \). This tells us that a score of 40 is 1 standard deviation below the mean. Likewise, you can do this for all other scores provided in the problem.

• Find the z-score for any value using the formula.
• A z-score helps connect a raw score to its relative position in the distribution.

By converting raw scores to z-scores, you can then use standard normal distribution tables to find probabilities associated with these z-scores.

The Role of Standard Deviation

Standard deviation is a measure of how spread out the values in a data set are around the mean. A low standard deviation means that the data points are close to the mean, while a high standard deviation means that data points are spread out over a wider range of values.

• In this exercise, the standard deviation is 10.
• It is used in the z-score formula to standardize raw scores.

Consider the z-score calculation: \( z = \frac{x - \text{mean}}{10} \). Here, the standard deviation helps reframe the score in a context that aligns with the distribution's characteristics. For example, a raw score of 60 with a mean of 50 and a standard deviation of 10 has a z-score of \( z = \frac{60 - 50}{10} = 1 \), indicating it is 1 standard deviation above the mean.

Interpreting the Mean

The mean, or average, is pivotal for understanding the distribution of data. It serves as the center point of a normal distribution. In this exercise, the mean is 50, making it the benchmark for computing z-scores and probabilities.

• The mean value is foundational for calculating z-scores.
• It helps us interpret the data's central tendency.

When you calculate a z-score, the mean acts as the reference point. For instance, if we look at a score of 47 in the distribution where the mean is 50, it is turned into a z-score by: \( z_{47} = \frac{47 - 50}{10} = -0.3 \). This tells us that 47 is 0.3 standard deviations below the mean of 50.