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1. Assume the readings on thermometers are normally distributed with a mean of 0 degrees and a standard deviation of 1.00 degrees C. Find the probablity P(z< -2.19 or z> 2.19, where z is the reading in degrees.

The probability of a temperature reading outside this range is

P = 0.028524

4. A simple random sample of FICO credit rating

scores is obtained, and the scores are listed below. As of this writing, the mean

FICO score was reported to be 676. Assuming the standard deviation of all

FICO scores is known to be 58.2, use a 0.05 significance level to test the claim

that these sample FICO scores come from a population with a mean equal to 676.

714, 751, 663, 790, 818, 779, 697, 836, 751, 834, 693, 800

What is the values of the test statistic? Z=

The p value is?=

Reject or fail HO? Is it sufficient yes or no?

The sample mean is 760.5. This is 84.5 higher than the population mean of 676.

With 12 samples, the standard error is 58.2 / sqrt(12) = 16.8.

The z-score for the sample mean is 84.5 / 16.8 = 5.0295.

The probability associated this z-score is 2 x 10^-7.

HO is that the population mean is 676. Our sample mean is sufficiently different from this, at the 0.05 confidence level, so we can reject H0.

5.Assume the readings on thermometers are normally distributed with a mean of 0 degrees and a standard deviation of 1.00. Find the probability?

P(-2.26 Pcritical, so we fail to reject H0.

Conclusion: the slot machine appears to be working.

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