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A variable is normally distributed with mean 68and standard deviation 10.

Part (a): Determine and interpret the quartiles of the variable.

Part (b): Obtain and interpret the 99thpercentile.

Part (c): Find the value that 85thof all possible values of the variable exceed.

Part (d): Find the two values that divide the area under the corresponding normal curve into a middle area of 0.90and two outside areas of 0.05. Interpret your answer.

Short Answer

Expert verified

Part (a): The first, second and third quartiles are 61.225,68and 74.745.

Part (b): 99%of all observations are smaller than 91.26.

Part (c): The value that 85%of all possible values of the variable exceeds is role="math" localid="1652456626971" 58.

Part (d): 90%of all observations are between51.55and84.449.

Step by step solution

01

Part (a) Step 1. Given information.

The given mean is 68and standard deviation is 10.

02

Part (a) Step 2. Determine the first and second quartiles of the variable.

The first quartile or 25th percentile of X is defined as PX≤x=0.25

PX-μσ≤x-μσ=0.25PZ≤x-6810=0.25x-6810=-0.6745x=61.255

The first quartile or 50th percentile of X is defined as PX≤x=0.5

PX-μσ≤x-μσ=0.5PZ≤x-6810=0.5x-6810=0x=68

03

Part (a) Step 3. Determine the third quartile of the variable.

The third quartile or 75th percentile of X is defined as PX≤x=0.75

PX-μσ≤x-μσ=0.75PZ≤x-6810=0.75x-6810=0.6745x=74.745

On interpreting, the first, second and third quartiles are 61.225,68and74.745.

04

Part (b) Step 1. Determine the 99th percentile of the variable.

The 99th percentile of X is defined as PX≤x=0.99

PX-μσ≤x-μσ=0.99PZ≤x-6810=0.99x-6810=2.326x=91.26

On interpreting, we get, 99%of all observations are smaller than91.26.

05

Part (c) Step 1. Determine the value that 85% of all possible values of the variable exceed.

The required value,

PX≥x=0.851-PX-μσ≤x-μσ=0.85PZ-x-6810=0.15x-6810=-1.036x=57.64

On interpreting, approximately 58 values of X exceeds 85%.

06

Part (d) Step 1. Find the two values.

The Z-score corresponding to the two outside areas of 0.05is given below,

z1=-1.645z2=1.645

Now, the first value is obtained,

x1=μ+z1σ=68+-1.64510=51.55

The second value is obtained,

x2=μ+z2σ=68+1.64510=84.449

On interpreting, we can say, 90%of all observations are between 51.551and 84.449.

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College-Math Success. Researchers S. Lesik and M. Mitchell explore the difficulty of predicting success in college-level mathematics in the article "The Investigation of Multiple Paths to Success in College-Level Mathematics" (fraternal of Applied Reacuwh in Hreher Eiturarion, Vol. 5. Issue 1. pP, 48-57). One of the variables explored as an indicator of success was the length of time since a college freshman has taken a mathematics course. The article reports that the mean length of time is 0.18 years with a standard deviation of 0.624 years. For college freshmen, let x represent the time, in years, since taking a math course.

A . What percentage of times are at least 0 years?

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