Final answer:
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.