find the indicated z score:.8907 0 z = 1.23. find the indicated z score:.6331 z 0 z = – 0.34.3669

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Find the indicated z score: Find the indicated z score: .8907 0 z = 1.23

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Page 1: Find the indicated z score:.8907 0 z = 1.23. Find the indicated z score:.6331 z 0 z = – 0.34.3669

Find the indicated z score:

Find the indicated z score:

.8907

0 z = 1.23

Page 2: Find the indicated z score:.8907 0 z = 1.23. Find the indicated z score:.6331 z 0 z = – 0.34.3669

Find the indicated z score:

.6331

z = – 0.34

.3669

Page 3: Find the indicated z score:.8907 0 z = 1.23. Find the indicated z score:.6331 z 0 z = – 0.34.3669

Find the indicated z score:

.3560

0 z = 1.06

.8560

Page 4: Find the indicated z score:.8907 0 z = 1.23. Find the indicated z score:.6331 z 0 z = – 0.34.3669

Find the indicated z score:

.4792

z = 0– 2.04

.0208

Page 5: Find the indicated z score:.8907 0 z = 1.23. Find the indicated z score:.6331 z 0 z = – 0.34.3669

Find the indicated z score:

0 z =

.4900

2.33

Page 6: Find the indicated z score:.8907 0 z = 1.23. Find the indicated z score:.6331 z 0 z = – 0.34.3669

Find the indicated z score:

z = 0

.005

– 2.575

Page 7: Find the indicated z score:.8907 0 z = 1.23. Find the indicated z score:.6331 z 0 z = – 0.34.3669

Find the indicated z score:

If area A + area B = .01, z = __________

A B

– z 0 z

2.575 or 2.58

= .005

Page 8: Find the indicated z score:.8907 0 z = 1.23. Find the indicated z score:.6331 z 0 z = – 0.34.3669

Application of Determining z Scores

The Verbal SAT test has a mean score of 500 and a standard deviation of 100.

Scores are normally distributed. A major university determines that it will accept

only students whose Verbal SAT scores are in the top 4%. What is the minimum

score that a student must earn to be accepted?

Page 9: Find the indicated z score:.8907 0 z = 1.23. Find the indicated z score:.6331 z 0 z = – 0.34.3669

...students whose Verbal SAT scores are in the top 4%.

Mean = 500, standard deviation = 100

= .04.9600

z = 1.75

The cut-off score is 1.75 standard deviations above the mean.

Page 10: Find the indicated z score:.8907 0 z = 1.23. Find the indicated z score:.6331 z 0 z = – 0.34.3669

Application of Determining z Scores

Mean = 500, standard deviation = 100

= .04.9600

z = 1.75

The cut-off score is 500 + 1.75(100) = 675.