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Ch 6.2 Application of Normal Distribution

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    15902
  • Ch 6.2 Using the Normal distribution

    When X is normally distributed with mean μ and standard deviation σ,  N( μ,σ) probability of  range of X can be represented by the area under normal  density curve \( y = \frac{e^{\frac{-1}{2}\cdot {(\frac{x-\mu}{\sigma})}^2}}{\sigma \sqrt{2 \pi}} \)   (no need to memorize.)

    Bell curve

     

     

     

    The same properties of bell shape density curve apply:

    Probability = area = relative frequency (percentage)

     a) P( X > μ ) = 0.5,  P(X < μ) = 0.5

     b) P(  a < X < b) = area between a and b under the density bell curve.

    A) Find Probability of X in Normal distribution

    Use online Normal distribution calculator to find prob.

    http://onlinestatbook.com/2/calculators/normal_dist.html

    Specify mean = μ, SD= σ

    -For left area or P(x < a) click below

    -For right area or P(x > a) click above

    -For area between two values a and b P( a < x < b), click between

    -For area outside of a and b, P(x <a or x > b), click outside

    -Click “Recalculate”

     

    B) Find X given probability in Normal Distribution

    Use Inverse Normal Calculator to find the cut-off given a left area, right area, between area or outside area of two tails.

    http://onlinestatbook.com/2/calculators/inverse_normal_dist.html

    Specify area, mean= μ, SD= σ

    select if area is below, above, between or outside.

    Click “Recalculate”

    Note: left area = bottom percentage = percentile.  Right area = top percent

     a) Find cutoff X for Top k percentage:

    Use Inverse online calculator, above

     b) Find cutoff X for Bottom k percentage or k percentile:  Use Inverse online calculator, below

     

    Ex1. Final exam of a standardized test is normally distributed with mean 63 and standard deviation 5.

    a) Find the probability that a randomly selected student scored more than 65 on the exam.

    normal bell curve

    Use Online Normal calculator.

    Mean = 63, SD =5,

    Click above, enter 65.

    Recalculate

    P( X > 65 ) = 0.3446.

     

    b) Find the probability that a score less than 85

    clipboard_ec6d740405ff9f49070e8de4d73d10563.png

    Use Online Normal calculator. Mean = 63, SD =5,

    Click below, enter 85.

    Recalculate

    P( X < 85 ) = 1

     

    c) Find the 90th percentile score.  ( top 10%)

    clipboard_ec6d740405ff9f49070e8de4d73d10563.png

    Use Inverse Normal Calculator, mean = 63, SD = 5

    Enter Area = 0.9, Click below

    The cut-off X is at the below box.

     P(X < __69.4___) = 0.9. The 90th percentile is 69.4.

     

    Ex2: Heights of men are normally distributed with mean of 68.6 in. and a standard deviation of 2.8 in.

    Find the probability that a randomly selected man has a height greater than 72 in.

    clipboard_ec6d740405ff9f49070e8de4d73d10563.png

     Use Online Normal calculator.

    Mean = 68.6, sd =2.8, Click above, enter 72.

    Recalculate

    P(X > 72) = 0.1123

     

    Ex3. Given pulse rate of women is normally distributed with a mean of 74 bpm and a standard deviation of 12.5 bpm. Find the pulse rates of lowest 5% and highest 5% of women.

    clipboard_ec6d740405ff9f49070e8de4d73d10563.png

    Total area = 5% + 5% =0.1

    Find cutoff pulse rate for lowest 5% and highest 5%. Use Inverse Normal calculator.

    Area = 0.10, Mean = 74, SD=12.5. Click Outside.

    Recalculate. 

    lowest 5% cutoff = 53.4 bpm highest 5%=94.6 bpm

    Ex4. A circus farmer grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and standard deviation of 0.24cm.

    a) Find the probability that a normally selected mandarin orange from his farm has a diameter larger than 6 cm.

    clipboard_ec6d740405ff9f49070e8de4d73d10563.png

    Use Online Normal calculator.

    Mean = 5.85, sd =0.24,

    Click above, enter 6.

    Recalculate

    P( X > 6 ) = 0.266

    b) Find the middle 20% diameter of mandarin oranges from his farm.

    clipboard_ec6d740405ff9f49070e8de4d73d10563.png

    Use Inverse Normal Calculator

    Area = 0.2, Mean = 5.85, sd=0.24

    Click Between.

      P( _5.79____< X < __5.91____) = 0.2

    Ex5. A TV has a life that is normally distributed with a mean of 6.5 years and a standard deviation of 2.3 years. If the company offers a warranty to replace any TV within 3 years. What percent of TV will need to be replaced?

    To find percent, use Normal online calculator.

    clipboard_ec6d740405ff9f49070e8de4d73d10563.png

    Mean = 6.5 , sd =2.3,

    Click below, enter 3.

    Recalculate

    P (X < 3) = 0.064 = 6.4%

     

     

     

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