Question

A variable is normally distributed with mean 9 and standard deviation 3.

a. Determine the quartiles of the variable
b. Obtain and interpret the 80th percentile
c. Find the value that 65% of all possible values of the variable exceed
d. Find the two values that divide the area under the corresponding normal curve into a middle area od 0.95 and two outside areas of .025. Then interpret the answer: These values enclose the area of the normal curve that is within __ standard deviations.

Answer 1

a. The quartiles of the variable with mean 9 and standard deviation 3 are Q1=7.025, Q2=9, and Q3=10.025. b. The 80th percentile is 11.535, meaning 80% of the data is below this value. c. The value that 65% of all possible values for the variable exceed is 7.845. d. the values that divide the area under the normal curve are 2.12 and 15.88, enclosing 1.96 standard deviations.

Step-by-step explanation:

a. The quartiles of a normal distribution can be found using z-scores. The first quartile (Q1) corresponds to a z-score of -0.675, the second quartile (Q2, which is the median) corresponds to a z-score of 0, and the third quartile (Q3) corresponds to a z-score of 0.675. To find the actual values, we can use the formula Q = mean + (z * standard deviation). So, Q1 = 9 + (-0.675 * 3) = 7.025, Q2 = 9 + (0 * 3) = 9, and Q3 = 9 + (0.675 * 3) = 10.025.

b. The 80th percentile is the value below which 80% of the data falls. To find this value, we can use the formula P = mean + (z * standard deviation), where P is the percentile as a decimal. So, P = 9 + (z * 3) = 9 + (0.845 * 3) = 11.535. This means that 80% of the data is below 11.535.

c. To find the value that 65% of all possible values exceed, we can use the formula P = mean + (z * standard deviation). Since we want the value that 65% of the data exceed, we subtract 65% from 100% to get 35%, which corresponds to a z-score of approximately -0.385. Therefore, the value is 9 + (-0.385 * 3) = 7.845.

d. To find the values that divide the area under the corresponding normal curve into a middle area of 0.95 and two outside areas of 0.025, we can use the z-score table. The z-score that corresponds to a middle area of 0.95 is approximately 1.96. Using the formula P = mean + (z * standard deviation), we can find the first value: 9 + (1.96 * 3) = 15.88. For the second value, we can use the formula P = mean - (z * standard deviation): 9 - (1.96 * 3) = 2.12. These values enclose the area of the normal curve that is within 1.96 standard deviations.

Answer 2

a. The first quartile (Q1) is 9.75 and the third quartile (Q3) is 10.25. b. The 80th percentile is 11.52. c. The value that 65% of all possible values exceed is 9.14. d. The values that divide the area under the normal curve into a middle area of 0.95 and two outside areas of 0.025 are 2.12 and 15.88.

Step-by-step explanation:

a. Quartiles:

The first quartile (Q1) can be found by subtracting 0.25 from the mean and multiplying the result by the standard deviation, then adding the mean. In this case, Q1 = (0.25 * 3) + 9 = 9.75. The third quartile (Q3) can be found by adding 0.25 to the mean and multiplying the result by the standard deviation, then adding the mean. In this case, Q3 = (0.25 * 3) + 9 = 10.25.

b. 80th percentile:

The 80th percentile can be interpreted as the value below which 80% of the data falls. To find the 80th percentile, we can use the formula: P = mean + (z * standard deviation), where P is the percentile, mean is the mean of the variable, z is the z-score corresponding to the percentile, and standard deviation is the standard deviation of the variable. Using the z-table, we find that the z-score corresponding to the 80th percentile is approximately 0.84. Plugging in the values, we get P = 9 + (0.84 * 3) = 11.52. Therefore, the 80th percentile is 11.52.

c. Value exceeding 65% of all possible values:

To find the value that 65% of all possible values exceed, we can use the formula: value = mean + (z * standard deviation), where value is the desired value, mean is the mean of the variable, z is the z-score corresponding to the percentage, and standard deviation is the standard deviation of the variable. Using the z-table, we find that the z-score corresponding to the 65th percentile is approximately 0.38. Plugging in the values, we get value = 9 + (0.38 * 3) = 9.14. Therefore, the value that 65% of all possible values exceed is 9.14.

d. Values dividing the area under the normal curve:

The two values that divide the area under the normal curve into a middle area of 0.95 and two outside areas of 0.025 can be found by subtracting and adding the z-scores corresponding to the outside areas from the mean, then multiplying the results by the standard deviation and adding the mean. Using the z-table, we find that the z-score corresponding to an area of 0.025 is approximately -1.96. Plugging in the values, we get value1 = (mean + (-1.96 * standard deviation)) = 9 + (-1.96 * 3) = 2.12 and value2 = (mean + (1.96 * standard deviation)) = 9 + (1.96 * 3) = 15.88. Therefore, the two values that divide the area under the normal curve are 2.12 and 15.88.