A variable is normally distributed with mean 16 and standard deviation 2.
B. The 90th percentile is the number that is $90 \%$ of the largest data value.
C. The 90th percentile is the number that divides the bottom $90 \%$ of the data from the top $10 \%$ of the data.
D. The 90th percentile is the number that divides the bottom 10\% of the data from the top 90\% of the data.
c. The value that $65 \%$ of all possible values of the variable exceed is $\square$ 15.22
(Round to two decimal places as needed.)
d. The two values that divide the area under the corresponding normal curve into a middle area of 0.95 and two outside area (Round to two decimal places as needed. Use ascending order.)
These values enclose the area of the normal curve that is within $\square$ standard deviations.
(Round to the nearest integer as needed.)
Answer (1)
The 90th percentile divides the bottom 90% of the data from the top 10%.
The value that 65% of all possible values exceed is calculated using the z-score for the 35th percentile: x = 16 + ( − 0.385 ) ( 2 ) = 15.23 .
The two values that divide the area into a middle area of 0.95 are calculated using z-scores of ± 1.96 : x 1 = 16 + ( − 1.96 ) ( 2 ) = 12.08 and x 2 = 16 + ( 1.96 ) ( 2 ) = 19.92 .
These values enclose an area within 2 standard deviations.
Explanation
Understand the problem and provided data We are given a normally distributed variable with a mean of 16 and a standard deviation of 2. We need to find the 90th percentile, the value that 65% of all possible values exceed, and the two values that divide the area under the normal curve into a middle area of 0.95.
Determine the correct definition of the 90th percentile The 90th percentile is the value below which 90% of the data falls. This is the same as saying that the 90th percentile divides the bottom 90% of the data from the top 10% of the data. So the correct answer is C.
Find the value exceeded by 65% of the data To find the value that 65% of all possible values exceed, we first need to find the corresponding z-score. This is the same as finding the 35th percentile. Using a z-table or calculator, the z-score for the 35th percentile is approximately -0.385. We can then use the formula x = μ + z σ to find the value.
Calculate the value Plugging in the values, we get x = 16 + ( − 0.385 ) ( 2 ) = 16 − 0.77 = 15.23 . Therefore, the value that 65% of all possible values of the variable exceed is 15.23.
Find the values that enclose the middle 95% of the data To find the two values that divide the area under the normal curve into a middle area of 0.95, we need to find the z-scores that correspond to the 2.5th percentile and the 97.5th percentile. These z-scores are -1.96 and 1.96, respectively. We can then use the formula x = μ + z σ to find the values.
Calculate the values Plugging in the values, we get x 0.025 = 16 + ( − 1.96 ) ( 2 ) = 16 − 3.92 = 12.08 and x 0.975 = 16 + ( 1.96 ) ( 2 ) = 16 + 3.92 = 19.92 . Therefore, the two values that divide the area under the normal curve into a middle area of 0.95 are 12.08 and 19.92.
Determine the number of standard deviations The z-scores that enclose the middle 95% of the data are ± 1.96 . Rounding to the nearest integer, we get 2 standard deviations.
Examples
Understanding percentiles and standard deviations is crucial in many real-world applications. For instance, in healthcare, these concepts help doctors interpret patient data, such as blood pressure or cholesterol levels, relative to the general population. In finance, they are used to assess investment risk and return. In education, standardized test scores are often reported in terms of percentiles, allowing educators to compare student performance across different schools or districts. These statistical tools provide a framework for making informed decisions based on data.
