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June 26th, 2014, 04:54 AM   #1
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Need help with Probability Problems

If anyone can help me solve these, I will forever be in you debt.

1-)The program office budgets \$30,000 per program review at the contractor’s site. Your concern for “end of year” spending drills is that you have budgeted enough for the reviews (i.e. you are only concerned if the actual costs are higher than the \$30,000 target). A sample of 16 trips yielded a mean of \$32,500 and a standard deviation of \$3,500. Test the budgeted amount at the 80% level of confidence. Select the correct answer out of each pair of choices. (Carry intermediate calculations to three decimal places.)
The tp is 1.341

The tp is 0.866


The tc is 11.429

The tc is 2.857


We would REJECT the null hypothesis

We would FAIL TO REJECT the null hypothesis


We would conclude that it is reasonable to USE THE \$30,000 BUDGET FIGURE

We would recommend REVISING THE BUDGET FIGURE


2-)We are estimating the spares requirement for a radar power supply. The power supply was designed with a mean (μ) life of 6500 hours. The standard deviation (σ) determined from testing is 750 hours. What is the likelihood that a power supply would fail between 7225 and 7670 hours?



A .0444 or 4.44%

B .3340 or 33.40%

C .1066 or 10.66%

D .4406 or 44.06%

3-) You are estimating the cost of engine overhauls. A sample of 49 repairs showed an average overhaul of 285 hours with a standard deviation of 60 hours. Calculate a 95% confidence interval for the average overhaul. (Carry intermediate calculations to three decimal places.)



A 281.61 ≤ μ ≤ 288.39

B 284.52 ≤ μ ≤ 285.48

C 268.20 ≤ μ ≤ 301.80

D 271.00 ≤ μ ≤ 299.00

Thanks in advance.

Last edited by skipjack; May 29th, 2015 at 02:42 PM.
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June 27th, 2014, 09:22 AM   #2
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3) Z-score for 95% CI will be between -1.96 and 1.96.
So, $\displaystyle 95 \text{%} CI = mean \pm 1.96 \times SD$.

2) $\displaystyle P(7225 \le x \le 7670 | \mu = 6500, \; \sigma = 750 ) \approx 0.1075$

Calculate Z-scores. $\displaystyle z_1 = \frac{7225-6500}{750} = 0.9666...... \; \& \; z_2 = \frac{7670-6500}{750} = 1.56$
FInd probabilities for $\displaystyle z_1$ and $\displaystyle z_2$, then subtract them.

1) Post it again, but the clear version.
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June 30th, 2014, 04:13 AM   #3
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Originally Posted by tahirimanov View Post
3) Z-score for 95% CI will be between -1.96 and 1.96.
So, $\displaystyle 95 \text{%} CI = mean \pm 1.96 \times SD$.

2) $\displaystyle P(7225 \le x \le 7670 | \mu = 6500, \; \sigma = 750 ) \approx 0.1075$

Calculate Z-scores. $\displaystyle z_1 = \frac{7225-6500}{750} = 0.9666...... \; \& \; z_2 = \frac{7670-6500}{750} = 1.56$
FInd probabilities for $\displaystyle z_1$ and $\displaystyle z_2$, then subtract them.

1) Post it again, but the clear version.
The program office budgets 30,000 per program review at the contractor’s site. Your concern for “end of year” spending drills is that you have budgeted enough for the reviews (i.e. you are only concerned if the actual costs are higher than the 30,000 target). A sample of 16 trips yielded a mean of 32,500 and a standard deviation of 3,500. Test the budgeted amount at the 80% level of confidence. Select the correct answer out of each pair of choices. (Carry intermediate calculations to three decimal places.)


The tp is 1.341

The tp is 0.866


The tc is 11.429

The tc is 2.857


We would REJECT the null hypothesis

We would FAIL TO REJECT the null hypothesis


We would conclude that it is reasonable to USE THE $30,000 BUDGET FIGURE

We would recommend REVISING THE BUDGET FIGURE
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August 24th, 2014, 12:34 PM   #4
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The program office budgets 30,000 per program review at the contractor’s site. Your concern for “end of year” spending drills is that you have budgeted enough for the reviews (i.e. you are only concerned if the actual costs are higher than the 30,000 target). A sample of 16 trips yielded a mean of 32,500 and a standard deviation of 3,500. Test the budgeted amount at the 80% level of confidence. Select the correct answer out of each pair of choices. (Carry intermediate calculations to three decimal places.)
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we are estimating the spares requirement for a radar power supply. the power supply was designed with a mean (μ) life of 6500 hours. the standard deviation (σ) determined from testing is 750 hours. what is the likelihood that a power supply would fail b
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We are estimating the spares requirement for a radar power supply. The power supply was designed with a mean (μ) life of 6500 hours. The standard deviation (σ) determined from testing is 750 hours. What is the likelihood that a power supply would fail b
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We are estimating the spares requirement for a radar power supply. The power supply was designed with a mean (μ) life of 6500 hours. The standard deviation (σ) determined from testing is 750 hours. What is the likelihood that a power supply would fail b
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We are estimating the spares requirement for a radar power supply. The power supply was designed with a mean (μ) life of 6500 hours. The standard deviation (σ) determined from testing is 750 hours. What is the likelihood that a power supply would fail b
,
we are estimating the spares requirement for a radar power supply. the power supply was designed with a mean (μ) life of 6500 hours. the standard deviation (σ) determined from testing is 750 hours. what is the likelihood that a power supply would fail b
,
we are estimating the spares requirement for a radar power supply. the power supply was designed with a mean (μ) life of 6500 hours. the standard deviation (σ) determined from testing is 750 hours. what is the likelihood that a power supply would fail b
,
We are estimating the spares requirement for a radar power supply. The power supply was designed with a mean (μ) life of 6500 hours. The standard deviation (σ) determined from testing is 750 hours. What is the likelihood that a power supply would fail b
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test the budgeted amount at the 80% level of confidence

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you are estimating the cost of engine overhauls. a sample of 99 repairs showed an

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the program office budgets $30,000 per program review at the contractor’s site. your concern for “end of year” spending drills is that you have budgeted enough for the reviews (i.e. you are only concerned if the actual costs are higher than the $30,

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