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 manics30 June 26th, 2014 03:54 AM

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.  tahirimanov June 27th, 2014 08:22 AM 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.  manics30 June 30th, 2014 03:13 AM Quote:  Originally Posted by tahirimanov (Post 198775) 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

 icarra1903 August 24th, 2014 11:34 AM

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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