Question

l. Armbrister Windows makes windows which have a production standard of an average of 0.375 inches thick with a standard deviation of 0.050 inches. A random sample of 64 windows yields an average of 0.382 inches. Calculate the probability of obtaining a random sample of 64 windows with an average of at least 0.382 inches. 2. The time it takes a mechanic to replace a brake pad is known to be normally distributed with a mean of 28 minutes and a standard deviation of 10 minutes. What size sample should be taken if the goal is to get the standard deviation of the sampling mean to be 3 minutes? 3. It is estimated that 82%ofall computer uses in Edina, Mn, make at least one purchase p month through the internet. In a random sample of 100 computer users from Edina, Mn, 74o f them made an internet purchase in the past month. Find the probability of finding no more than 74 computer users from Edina, MN making an internet purchase from a sample of 100. 4. Marketing research indicates that 37% of all customers at a nationwide pizza chain are college students. A random sample of 50 customers at a local franchise yields 15 college student customers. Find the probability that a random sample of 50 customers would yield between 14 and 20 college student customers.

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Answer 1

Solution (1):

     We are given that : Armbrister windows makes windows which have a production standard of an average of 0.375 inches thick with a standard deviation of 0.050 inches.

Thus Mean = $\mu = 0.375$ and standard deviation = $\sigma = 0.050$

A random sample of 64 windows yields an average of 0.382 inches.
We have to find the probability that : a random sample of 64 windows with an average of at least 0.382 inches.

That is we have to find : $P(\bar{x}\geq 0.382)=.......?$

Since sample size n = 64 > 30 , we can assume large sample and hence using Central limit theorem , sampling distribution of sample mean has an approximate Normal distribution with mean of sample means = $\mu_{\bar{x}} = \mu = 0.375$ and standard deviation of sample means = $\sigma_{\bar{x}} = \sigma / \sqrt{n} = 0.050/ \sqrt{64} = 0.050 / 8 = 0.00625$

Thus to find : $P(\bar{x}\geq 0.382)=.......?$ find z score .

$z= \frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}$

$z= \frac{0.382-0.375}{0.00625}$

$z= \frac{0.007 }{0.00625}$

Thus we get :

$P(\bar{x}\geq 0.382)=P(z\geq 1.12)$

$P(\bar{x}\geq 0.382)=1-P(z\leq 1.12)$

Look in z table for z = 1.1 and 0.02 and find corresponding area.

100 oil 02 .05 .06 07 08 .09 0.0 .5000 .5040 . 0.1 .5398 .5438 0.2 .5793 5 5080 .5120 5160 5199 5239 5279 5319 5359 5517 5557 5596 .5636 .5675 .5714 5753 .5910 .5948 .59876026 .6064 .6103 6141 5832 .5871 0.6 .7257 .7291-4 .7357 .7389 .7422 .7454 .7486 .7517 7549 0.7 7580 .7611 0.8 7881 .7910.7989 0.9 .8159 .8186 7612 7673 .7704 7734 7764 7794 .7823 .7852 .822 2 .8238 8264 8289 .8315 .8340 .8365 .8389 8577 8599 .8621 .8708 8729 8749 8770 8790 .8810 8830 8413 .8438 8481 . 8485 .8508 .8531 1.1 8686

P( Z < 1.12 ) = 0.8686

Thus

$P(\bar{x}\geq 0.382)=1-P(z\leq 1.12)$

$P(\bar{x}\geq 0.382)=1-0.8686$

$P(\bar{x}\geq 0.382)=0.1314$

Thus the probability that a random sample of 64 windows with an average of at least 0.382 inches is 0.1314 .