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 If E and F are two independent events with P(E)=.2 and P(F)=.4 find:  1) P(E|F)  2) P(F|E)  3) P(E∩F)  4) P(EUF)  1).22..4 3).083).52 WARM UP:

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Presentation on theme: " If E and F are two independent events with P(E)=.2 and P(F)=.4 find:  1) P(E|F)  2) P(F|E)  3) P(E∩F)  4) P(EUF)  1).22..4 3).083).52 WARM UP:"— Presentation transcript:

1  If E and F are two independent events with P(E)=.2 and P(F)=.4 find:  1) P(E|F)  2) P(F|E)  3) P(E∩F)  4) P(EUF)  1).22..4 3).083).52 WARM UP:

2 SWBAT compute a z-score, use the standard normal curve, and approximate the binomial distribution. 9.6 THE NORMAL DISTRIBUTION

3  Bell shaped and symmetric with respect to mean.  Mean, median and mode are equal.  Area enclosed by x-axis and curve’s = 1sq unit.  Probability that an outcome of a normally distributed curve is between b and a equals the area associated w the curve from x=a to x = b.  Standard deviation plays a major role in describing the area under the curve. PROPERTIES OF NORMAL DISTRIBUTION:

4 STANDARD DEVIATION RELATED TO AREA UNDER A NORMAL CURVE: µ-3σ µ-2σ µ-σµ+σ µ+2σ µ+3σ µ 2.14% 13.59% 34.135%

5 Z-SCORE: (A GROUP MEAN) Z- score transforms any normal value with a mean  and standard deviation  to a normal value with mean 0 and standard deviation 1.

6 ON A TEST, 80 IS THE MEAN AND 6 IS THE STANDARD DEVIATION.

7  See the Z-Score Table. FIND THE AREA, THAT IS FIND THE PROPORTION OF CASES INCLUDED BETWEEN THE GIVEN VALUES: 1.2 1.940

8  At 1.2: 0.3849  At 1.94: 0.4738  So we will subtract the two since.3849 is the area from 0 to 1.2 and.4738 is the area from 0 to 1.94. .4738 -.3849 =.0889 FIND TWO VALUES: SCORE AT 1.2 AND THE SCORE AT 1.94.

9 SCORES ON A TEST ARE NORMALLY DISTRIBUTED WITH A MEAN OF 75 AND A STANDARD DEVIATION OF 8. WHAT IS THE PROBABILITY THAT A TEST CHOSEN AT RANDOM IS BETWEEN 80 AND 90?

10 *Keep your Z- score chart to use on your test!! HW WS 9.6: 1,3, 5A-C, 7A-C, 9,11,13, 15A-C, 19


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