Satyagopal Mandal
Department of Mathematics
University of Kansas
Office: 624 Snow Hall  Phone: 785-864-5180
  • e-mail: mandal@math.ukans.edu
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    Normal Distribution 2

     

    We have discussed both discrete and continuous random variables. We have to note here that a normal random variable X is a continuous random variable.

    For a normal random variable X, we only know 68-95-99 rule to compute probability that X is within 1, 2 or 3 standard deviation s from the mean m. We will learn how to compute the probability that X is within any two given numbers.

    First, we can rephrase the definition of the normal distribution as follows:

    Suppose that a random variable X (or a data set) has normal distribution with mean m and standard deviation s. Then the relative frequency histogram of the data has a perfect bell shape as follows.




    Picture 19

    Now we have the following:

    1. As before, the graph is symmetric around the vertical line through the mean m.

    2. The total area under the normal curve and above the horizontal axis (x-axis) is one. So, on two sides of the line of symmetry the area under the graph is half.

    3. The Probability rule: The probability that X is between two given number a and b ( with a < b) is equal to the area under the normal curve, above the x-axis and between the vertical lines x = a and x = b. So,

      4. P(a < X < b) = P(a < X < b ) =

      P(a < X < b) = P(a < X < b) =

      the area under the normal curve, above the x-axis and between the vertical lines x = a and x = b.

       


      Picture 20

      Remark: Note that this formula says the probability that X is between a or b. The formula does not care if we include one or both a and b or not.

       

    4. The formula also works if a is negative infinity and b is infinity. That means,

      5. P( X < b) = P( X <b) =

      the area under the normal curve, above the x-axis and on the left side of the vertical line x = b.

       


      Picture 21

       

      and

      P(a < X) = P(a < X) =

      the area under the normal curve, above the x-axis and on the right side of x = a.

       

       


      Picture 22

       

    5. It follows that for any number a, we have P (X=a) = 0. This is very much in contrast with the case of discrete variables.

     

    The Z-Table

  • The Z-Table

     

    (The z-table is also available separately.)

    The Z-Table

    Recall that the normal random variable with mean zero and standard deviation one is called standard random variable (denoted by Z) .As we have said before we will give a table for the standard normal random variable Z. The following table gives the P(Z < z) for different values of z.

     


    Picture 23

     

    z

    -3.0

    -2.9

    -2.8

    -2.7

    -2.6

    P(Z<z)

    0.0013

    0.0019

    0.0026

    0.0035

    0.0047

     

     

    z

    -2.5

    -2.4

    -2.3

    -2.2

    -2.1

    P(Z<z)

    0.0062

    0.0082

    0.0107

    0.0139

    0.0179

     

     

    z

    -2.0

    -1.9

    -1.8

    -1.7

    -1.6

    P(Z<z)

    0.0228

    0.0287

    0.0359

    0.0446

    0.0548

     

     

    z

    -1.5

    -1.4

    -1.3

    -1.2

    -1.1

    P(Z<z)

    0.0668

    0.0808

    0.0968

    0.01151

    0.1357

     

    z

    -1.0

    -0.9

    -0.8

    -0.7

    -0.6

    P(Z<z)

    0.1587

    0.1841

    0.2119

    0.2420

    0.2743

     

     

    z

    -0.5

    -0.4

    -0.3

    -0.2

    -0.1

    P(Z<z)

    0.3085

    0.3446

    0.3821

    0.4207

    0.4602

     

     

     

    z

    0.0

    0.1

    0.2

    0.3

    0.4

    P(Z<z)

    0.5000

    0.5398

    0.5793

    0.6179

    0.6554

     

     

    z

    0.5

    0.6

    0.7

    0.8

    0.9

    P(Z<z)

    0.6915

    0.7257

    0.7580

    0.7881

    0.8159

     

    z

    1.0

    1.1

    1.2

    1.3

    1.4

    P(Z<z)

    0.8413

    0.8643

    0.8849

    0.9032

    0.9192

     

    z

    1.5

    1.6

    1.7

    1.8

    1.9

    P(Z<z)

    0.9332

    0.9452

    0.9554

    0.9641

    0.9713

     

    z

    2.0

    2.1

    2.2

    2.3

    2.4

    P(Z<z)

    0.9772

    0.9821

    0.9861

    0.9893

    0.9918

     

    z

    2.5

    2.6

    2.7

    2.8

    2.9

    P(Z<z)

    0.9938

    0.9953

    0.9965

    0.9974

    0.9981

     

    The Final Formula:

    P(a < X < b) =

    P((a-m)/s <(X- m)/ s < (b- m)/ s) =

    P((a- m)/ s < Z < (b- m)/ s) = The difference of the two numbers to be read from the z-table.

     

     

     

     

    Problems on Normal Distribution

     

    Ex. 1: The length X of life of some light bulbs produced in a factory is normally distributed with mean m= 8640 hours and standard deviation s = 1440 hours.

    1) Find the probability of the event E that a bulb will last less than 5040 hours.

    2) Find the probability of the event F that a bulb will last between 5040 hours and 8640 hours.

     

    Solution 1) We have P(E) = P( X < 5040) =

    P((X-m)/s < (5040-m)/s) =

    P(Z < (5040-8640)/1440) =

    P(Z < -2.5) = 0.0062 (This is from the table).

     

    Solution 2) We have P(F) = P( 5040 < X < 8640) =

    P((5040 -m)/s < (X-m)/s < (8640-m)/s ) =

    P((5040 -8640)/1440 < Z < (8640-8640)/1440 ) =

    P(-2.5 < Z < 0) =

    P(Z < 0) - P(Z < -2.5) =

    0.5 - 0.0062 = 0.4938.

     

     

    Ex. 2: The diameter X of a bolt produced by a machine in normally distributed with a mean m = 0.82 cm and a standard deviation s =0.004 cm. What percent of the bolts will meet the specification that they be between 0.8144 cm and 0.8268 cm in diameter.

     

    Solution: Here we are asked to find P(0.8144 < X < 0.8268). (They want the answer in percent.) We have

    P( 0.8144 < X < 0.8268) =

    P((0.8144 -m)/s < (X-m)/s < (0.8268-m)/s ) =

    P((0.8144 -0.82)/0.004 < Z < (0.8268-0.82)/0.004 ) =

    P(-1.4 < Z < 1.7) =

    P(Z < 1.7) - P(Z < -1.4) = 0.9554-0.0808 = 0.8746. So, the answer is 87.46 percent.

     

     

    Ex. 3: The height X of an adult male is known to be normally distributed with mean m = 69 inches and a standard deviation s = 2.5 inches. What percent of the adult male population is above 72 inches?

     

    Solution: Here we are asked to find P(72 < X). (They want the answer in percent.) We have

    P(72 < X) =

    P((72 -m)/s < (X-m)/s ) =

    P((72 -69)/2.5 < Z ) =

    P(1.2 < Z) = 1 - P(Z < 1.2) =1- 0.8849 = 0.1151. So, the answer is 11.51 percent.

     

     

    Ex. 4: The diameter X of a lead bolt produced by a machine has normal distribution with mean m = 2 inches and standard deviation s = 0.05 inches. What percent of the bolts will be between 1.88 inches and 2.11 inches?

     

    Solution: Here we are asked to find P(1.88 < X < 2.11). (They want the answer in percent.) We have

    = P(1.88< X < 2.11) =

    P((1.88 -m)/s < (X-m)/s < (2.11-m)/s ) =

    P((1.88-2)/0.05 < Z < (2.11-2)/0.05 ) =

    P(-2.4 < Z < 2.2) =

    P(Z < 2.2) - P(Z < -2.4) =0.9861 - 0.0082 = 0.9779. So, the answer is 97.79 percent.

     

    Ex. 5: Suppose the systolic blood pressure of an adult male is normally distributed with mean m = 138 mm and standard deviation s = 10 mm. What percent of adult male population has systolic blood pressure above 157 mm?

     

    Ex. 6: The annual expenditure X of a student is approximately normally distributed with mean m = 11,000 dollars and standard deviation s = 1500 dollars. Approximately what percent of students spend less than 10,000 dollars?

     

    Ex. 7: Suppose the annual production X of milk per cow is normally distributed with m = 5500 liters and standard deviation s = 150 liters. What percent of cows have annual yield less than 5155 liters?

     

    Ex. 8: The amount eatable oil X produced by a machine in a day is normally distributed with m = 130 liters and standard deviation s = 25 liters. What is the probability that a machine will produce between 120 liters and 150 liters on a day?