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# Normal Distribution & Critical Z-Value

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1.The net profit of an investment is normally distributed with a mean of \$10,000 and a standard deviation of \$5,000. The probability that the investor's net profit will be between \$12,000 and \$15,000 is _____________.
a) 0.8413
b) 0.1859
c) 0.4967
d) 0.3413

2. When the rod shearing process at Stockton Steel is "in control" it produces rods with a mean length of 120 inches. Periodically, quality control inspectors select a random sample of 36 rods. If the mean length of sampled rods is too long or too short, the shearing process is shut down. Sarah Shum, Director of Quality Programs, chose a 0.05 level of significance for this test. The critical z values are _________.
a) -1.75 and 1.75
b) -1.96 and 1.96
c) -1.645 and 1.645
d) -2.33 and 2.33

https://brainmass.com/statistics/probability/normal-distribution-critical-z-value-101448

#### Solution Preview

1.The net profit of an investment is normally distributed with a mean of \$10,000 and a standard deviation of \$5,000. The probability that the investor's net profit will be between \$12,000 and \$15,000 is _____________.
a) 0.8413
b) 0.1859
c) 0.4967
d) 0.3413

Mean=M = \$10,000
Standard deviation =s= \$5,000
x1= \$12,000
x2= \$15,000
z1=(x1-M )/s= 0.4 ...

#### Solution Summary

Answers multiple choice questions on Normal Distribution & Critical Z-Value

\$2.19

## Find Critical Z Value Assuming Normal Distribution

Please help me solve the following, I need the steps so that I can do them myself. Let me know how many credits will be required.

1.)Find the critical z value, assume that the normal distribution applies.
alpha = 0.005; H1 is p not equal 0.20

2.)In 1990, 5.8% of job applicants who were tested for drugs failed the test. At the 0.01 significance level, test the claim that the failure rate is now lower if a simple random sample of 1520 current job applicants results in 58 failures. Does the result suggest that fewer job applicants now use drugs?

3.) Claim: The mean life span of desktop PC's is less than 7 years. Sample data: n=21, mean of the value = 6.8 years, s = 2.4 years. The significance level is alpha = 0.05.
Find the null and alternative hypothesis, test statistic, critical value, and state the final conclusion. Assume that a simple random sample has been selected from a normally distributed population

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