In a large section of a statistics class, the points for the final exam are normally distributed, with a mean of 73 and a standard deviation of 8. Grades are assigned such that the top 10% receive A's, the next 20% received B's, the middle 40% receive C's, the next 20% receive D's, and the bottom 10% receive F's. Find the lowest score on the final exam that would qualify a student for an A, a B, a C, and a D.The lowest score that would qualify a student for an A is nothing.(Round up to the nearest integer as needed.)The lowest score that would qualify a student for a B is nothing.(Round up to the nearest integer as needed.)The lowest score that would qualify a student for a C is nothing.(Round up to the nearest integer as needed.)The lowest score that would qualify a student for a D is nothing.(Round up to the nearest integer as needed.) This image displays a standard normal distribution table, commonly known as a Z-table. The Z-table provides the cumulative probability of a standard normal random variable being less than a given value. It is commonly used in statistical analyses.### Description of the Table:- **Columns and Rows:** - The leftmost column represents the Z-score values, covering a range from -3.4 to -1.3 in this portion of the table. - The top row displays decimal values ranging from 0.00 to 0.09.- **Values:** - Each cell in the table provides the cumulative probability that a standard normal variable is less than the associated Z-score. - For example, the top-left cell (column 0.00, row -3.4) corresponds to a probability of 0.0002, meaning there is a 0.02% likelihood that a Z-score is less than -3.4.### Usage:To find the probability of a Z-score:1. Locate the Z-score's first decimal value in the leftmost column.2. Identify the second decimal value in the top row.3. Find the intersection of the row and column for the cumulative probability.This table is an essential tool in statistics for calculating probabilities and understanding the distribution of datasets. This image appears to be a Z-table, which is used in statistics to find the probability of a given score occurring within a standard normal distribution. The Z-table typically provides the area (probability) to the left of a given Z-score on a standard normal curve.### Understanding the Table:- **Rows & Columns**: The table is organized with Z-scores listed along the leftmost column and across the top row. Each Z-score is a combination of a row and column value.- **Values**: Each cell within the table represents the cumulative probability associated with its corresponding Z-score.### Example:To find the cumulative probability for a Z-score of 0.53:1. **Row**: Locate the row corresponding to 0.5 (first digit and first decimal place of the Z-score).2. **Column**: Move to the column for 0.03 (second decimal place of the Z-score).3. **Intersection**: The value at this intersection gives the cumulative probability for Z = 0.53. This Z-table is an essential tool for statistics students and professionals, providing quick reference to calculate probabilities for normal distributions.

Name: In a large section of a statistics class, the points for the final ex
Uploaded: 2024-03-20T07:07:12.000Z
Channel: Bartleby
Description: In a large section of a statistics class, the points for the final exam are normally distributed, with a mean of 73 and a standard deviation of 8. Grades are assigned such that the top 10% receive A's, the next 20% received B's, the middle 40%

Solution: Let X be the random variable defines as the points for the final exam. X is normally distributed, with a mean of 73 and a standard deviation of 8. X ~ N( μ=73, σ=8 ) It is given that the grades are assigned such that the top 10% receive A's, the next 20% received B's, the middle 40% receive C's, the next 20% receive D's, and the bottom 10% receive F's. The lowest score on the final exam that would qualify a student for an A is required. A student would qualify with grade A if he is in top 10% students. That means, right tail area is given to be 10% = 0.10. Left tail area = 1-Rigth tail area = 1-0.10 = 0.90 Look for 0.90 or its nearest value in the body of the table. The nearest value of 0.9000 in the above table is 0.8997. The row heading corresponding to 0.8997 is 1.2 and column heading is 0.08. Critical value = Row heading + Column heading = 1.2+0.08 =1.28 Thus, critical value: z = 1.28 Rounding it to the nearest integer, it becomes 83. Thus, the lowest score on the final exam that would qualify a student for an A is 83. The lowest score that would qualify a student for a B is required. It is given that the grades are assigned such that the top 10% receive A's, the next 20% received B's. That means, the value for which the area to the right of it is 30% is required. Right tail area = 30% = 0.30 Left tail area = 1- right tail area = 1-0.30 = 0.70 Look for 0.7000 or its nearest value in the body of the table. The nearest value of 0.7000 in the above table is 0.6985. The row heading corresponding to 0.6985 is 0.5 and column heading is 0.02. Critical value = Row heading + Column heading = 0.5+0.02 = 0.52 Thus, critical value: z = 0.52 Rounding it to the nearest integer, it becomes 77. Thus, the lowest score on the final exam that would qualify a student for a B is 77. The lowest score that would qualify a student for a C is required. It is given that the middle 40% receive C's. That means, middle area is 0.40. That means, the value for which the area to the right of it is (40%+20%+10%=) 70% is required. Right tail area = 70% = 0.70 Left tail area = 1- right tail area = 1-0.70 = 0.30 Look for 0.3000 or its nearest value in the body of the table. The nearest value of 0.3000 in the above table is 0.3015. The row heading corresponding to 0.3015 is -0.5 and column heading is 0.02. Critical value = Row heading + Column heading = -0.5+0.02 = -0.52 Thus, critical value: z = -0.52 Rounding it to the nearest integer, it becomes 69. Thus, the lowest score on the final exam that would qualify a student for a C is 69. The lowest score that would qualify a student for a D is required. It is given that the middle 40% receive C's, the next 20% receive D's. So, bottom 10% to bottom 30% would receive grade D. We want the lowest score that would qualify a student for a D. SO, we will find the score for which area to the left of it is 0.10. Thus, left tail area = 0.10 Look for 0.1000 or its nearest value in the body of the table. The nearest value of 0.1000 in the above table is 0.1003. The row heading corresponding to 0.1003 is -1.2 and column heading is 0.08. Critical value = Row heading + Column heading = -1.2+0.08 = -1.28 Thus, critical value: z = -1.28 Rounding it to the nearest integer, it becomes 63. Thus, the lowest score on the final exam that would qualify a student for a D is 63.