8-Determinig-Probdgfdghgfdsabilities.pptx

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1. DETERMINING PROBABILITIES DOMINIC DALTON L. CALING Statistics and Probability | Grade 11

2. LESSON OBJECTIVES At the end of this lesson, you are expected to: find areas between paired z-scores; find probabilities for the standard normal random variable z; and express areas under the normal curve using probability notation.

3. Pre-Assessment

4. Lesson Introduction S tandard normal distribution is a normal distribution with and . A random variable with a standard normal distribution, denoted by X, is called a standard normal random variable. P robabilities associated with the standard normal random variables can be shown as areas under the standard normal curve.  

5. Discussion Points Probability Notations Under the Normal Curve The following notations for a random variable are used in our various solutions concerning the normal curve. P(a< z < b) denotes the probability that the z-score is between a and b . P(z > a) denotes the probability that the z-score is greater than a . P(z < a) denotes the probability that the z-score is less than a . where a and b are z-score values.

6. Discussion Points T o find the area of the region between z = 1 and z = 2, we subtract .3413 from .4772 resulting in .1359. It is graphically shown below.

7. Discussion Points T he regions under the normal curve in terms of percent, the graph of the distribution would look like this:

8. Discussion Points

9. EXAMPLE 1 Find the proportion of the area above z = –1. STEPS SOLUTION Draw a normal curve. Locate the z-value. Draw a line through the z-value Shade the required region. Consult the z-Table and find the area that corresponds to z = –1. z = –1 corresponds to an area of .3413 Examine the graph and use probability notation to form an equation showing the appropriate operation to get the required area. The graph suggests addition. The required area is equal to 0.3413 + 0.5 = 0.8413. That is, P(z > –1) = 0.3413 + 0.5 = 0.8413 Make a statement indicating the required area. The proportion of the area above is .8413 .

10. EXAMPLE 2 STEPS SOLUTION Draw a normal curve. Locate the z-value. Draw a line through the z-value Shade the required region. Consult the z-Table and find the area that corresponds to z = –1. 5 z = 1.5 corresponds to the area 0.4332 Examine the graph and use probability notation to form an equation showing the appropriate operation to get the required area. The graph suggests subtraction. The required area is equal to 0.5 000 – 0.4332 = 0.0668 That is, P( z < –1.5) = 0.5 000 – 0.4332 = 0.0668 Make a statement indicating the required area. The proportion of the area to the left of z = –1.5 is 0.0668 . Find the area to the left of z = –1.5.

11. EXAMPLE 3 STEPS SOLUTION Draw a normal curve. Locate the z-value. Draw a line through the z-value Shade the required region. Consult the z-Table and find the area that corresponds to z = –1. 5 z = –2 corresponds to .4772 z = –1.5 corresponds to .4332. Examine the graph and use probability notation to form an equation showing the appropriate operation to get the required area. The graph suggests subtraction. The required area is equal to 0.4772 – 0.4332 That is, P (–2 < z < –1.5) = 0.4772 – 0.4332 = 0.0440 Make a statement indicating the required area. The required area between z = –2 and z = –1.5 is 0.0440 . Find the area between z = –2 and z = –1.5.

12. EXERCISES Find the area between z = –1.32 and z = 2.37. Complete the table below. z = -1.32 corresponds to 0.4066 z = 2.37 corresponds to 0.4911 The graph suggests addition. The required area is equal to 0.4066 + 0.4911 That is, P (–1.32 < z < 2.37) = 0.4066 + 0.4911 = 0.8977 The required area between z = –1.32 and z = 2.37 is 0.8977 .

13. EXERCISES Determine each of the following areas and show these graphically. Use probability notation in your final answer. above z = 1.46 b elow z = –0.58 between z = – 0. 78 and z = –1.95 between z = 0.76 and z = 2.88 between z = –0.92 and z = 1.75

14. EXERCISES 1. Find the area above z = 1.46. z = 1.46 corresponds to 0.4279 The graph suggests subtraction. The required area is equal to 0.5000 – 0.4279 That is, P (z > 1.46) = 0.5000 – 0.4279 = 0.0721 The required area above z = 1.46 is 0.0721 . z = 1.46.

15. EXERCISES 2. Find the area below z = – 0.58. z = – 0.58 corresponds to 0.2190 The graph suggests subtraction. The required area is equal to 0.5000 – 0.2190 That is, P (z < -0.58) = 0.5000 – 0.2190 = 0.2810 The required area below z = – 1.46 is 0.2810 . z = – 0.58.

16. EXERCISES 3. Find the area between z = – 0.78 and z = –1.95. z = – 0.78 corresponds to 0.2823 z = – 1.95 corresponds to 0.4744 The graph suggests subtraction. The required area is equal to 0.4744 – 0.2823. That is, P (-0.78 < z < -1.95) = 0.4744 – 0.2823 = 0.1921. The required area between z = –0.78 and z = –1.95 is 0.1921. z = – 0.78 and z = – 1.95.

17. EXERCISES 4. Find the area between z = 0.76 and z = 2.88. z = 0.76 corresponds to 0.2764 z = 2.88 corresponds to 0.4980 The graph suggests subtraction. The required area is equal to 0.4980 – 0.2764. That is, P (0.76 < z < 2 .88) = 0.4980 – 0.2764 = 0.2216. The required area between z = 0.76 and z = 2.88 is 0.2216. z = 0.76 and z = 2.88.

18. EXERCISES 5. Find the area between z = – 0.92 and z = 1.75. z = – 0.92 corresponds to 0.3212 z = – 1.75 corresponds to 0.4599 The graph suggests addition. The required area is equal to 0.3212 + 0.4599. That is, P (-0.92 < z < 1.75) = 0.3212 + 0.4599 = 0.7811. The required area between z = –0.92 and z = 1.75 is 0.7811. z = – 0.92 and z = 1.75.

19. SUMMARY Steps in Determining Areas Under the Normal Curve Use a cardboard model to draw a normal curve. Locate the given z-value or values at the base line. Draw a vertical line through these values. Shade the required region. Find models, if any. Consult the z-Table to find the areas that correspond to the given z-value or values. Examine the graph and use probability notation to form an equation showing an appropriate operation to get the required area. Make a statement indicating the required area.

20. THANK YOU!