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dalvyx [7]
2 years ago
8

A random sample of 85 group leaders, supervisors, and similar personnel at General Motors revealed that, on average, they spent

6.5 years in a particular job before being promoted. The standard deviation of the sample was 1.7 years. Construct a 95% confidence interval.
Mathematics
1 answer:
andreyandreev [35.5K]2 years ago
4 0

Answer:

Step-by-step explanation:

From the information given,

Number of personnel sampled, n = 85

Mean or average = 6.5

Standard deviation of the sample = 1.7

We want to determine the confidence interval for the mean number of years that personnel spent in a particular job before being promoted.

For a 95% confidence interval, the confidence level is 1.96. This is the z value and it is determined from the normal distribution table. We will apply the following formula to determine the confidence interval.

z×standard deviation/√n

= 1.96 × 6.5/√85

= 1.38

The confidence interval for the mean number of years spent before promotion is

The lower end of the interval is 6.5 - 1.38 = 5.12 years

The upper end is 6.5 + 1.38 = 7.88 years

Therefore, with 95% confidence interval, the mean number of years spent before being promoted is between 5.12 years and 7.88 years

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Comment
This is an area problem. The key words are 120 square feet and 12 feet longer.
And of course width is a key word when you are reading this.

Formula
Area = L * W

Givens
W = W
L = W + 12

Substitute and Solve
Area = L* W
120 = W*(W + 12)
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This does not factor easily. I would have thought that a graph might help but not if the dimension has to be to the nearest 1/100 of a foot. The only thing we can do is use the quadratic formula.

a = 1
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w = [ -b +/- sqrt(b^2 - 4ac) ]/(2a)
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w = [-12 +/- sqrt(144 - (-480)]/2
w = [-12 +/- sqrt(624)] / 2
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w = (12.979992) / 2
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L = w + 12
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check
Area = L * W
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Answer
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W = 6.489995 = 6.49 feet

Note: in the check if you round first to the answer, LW = 120.0001 when you find the area for the check. Kind of strange how that nearest 1/100th makes a difference.
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