1. ## Questions I Missed

When sampling is done for the proportion of defective items in a large shipment, where the population proportion is 0.07, and the sample size is 200, what is the probability that the sample proportion will be at least 0.09?
(a) 0.3665 (b) 0.1333 (c) 0.8667 (d) 0.6335

When sampling is from a population with standard deviation =35, using a sample of size n=125, what is the probability that will be at least 6 units away from the population mean ?
(a) 0.9448 (b) 0.0274 (c) 0.4726 (d) 0.0552

A large lot of parts is 20% defective. A random sample of 15 parts is to be selected. Let (p with ^) denote the sample proportion of defective parts. Use a binomial distribution to determine P( (p with ^) = 4/15)
(a) 0 (b) 0.648 (c) 0.250 (d) 0.188 (e) none of these

The amount of pollutants released from a smoke stack in a day is a normally distributed with mean 3 ounces and standard deviation 0.6 ounces.

What is the probability that the average release over 9 days is within 0.34 ounces of the population mean?
(a) 0.9108 (b) 0.4554 (c) 0.9554 (d) 0.0892

Could anyone give me some hints on how to figure out the correct answers? I'm using a TI-83 calculator and use a lot of the Distribution commands / functions / whatever you want to call them.

2. ## Re: Questions I Missed

Originally Posted by randomness
A large lot of parts is 20% defective. A random sample of 15 parts is to be selected. Let (p with ^) denote the sample proportion of defective parts. Use a binomial distribution to determine P( (p with ^) = 4/15)
(a) 0 (b) 0.648 (c) 0.250 (d) 0.188 (e) none of these
I think I figured this one out.

n = 15
p = .20
x = 4 / 15 * 15 = 4

so... binompdf(15,.2,4) = .188

Still lost on the others.

3. ## Re: Questions I Missed

Originally Posted by randomness
When sampling is from a population with standard deviation =35, using a sample of size n=125, what is the probability that will be at least 6 units away from the population mean ?
(a) 0.9448 (b) 0.0274 (c) 0.4726 (d) 0.0552
Not sure if this is right, but..

Z < (6 / (35 / sqrt(125))
Z < 1.917
p = 1 - .972381 = .0276

Seems a bit off, I don't think I did it right. I know that normally the formula is..

Z < (X(bar) - mean / (std / sqrt(n)))

The amount of pollutants released from a smoke stack in a day is a normally distributed with mean 3 ounces and standard deviation 0.6 ounces.

What is the probability that the average release over 9 days is within 0.34 ounces of the population mean?
(a) 0.9108 (b) 0.4554 (c) 0.9554 (d) 0.0892
I thought maybe this one would be,

Over 9 days..

mean = 9 * 3 = 27

normalcdf(27-.34,27+.34,27,.6) = .429

so I thought maybe then STD had to be taken in account for time also, so I did STD * sqrt(9) and got 1.8

normalcdf(27-.34,27+.34,27,.6) = .150

Looks like I'm really far off.

4. ## Re: Questions I Missed

Originally Posted by randomness
When sampling is done for the proportion of defective items in a large shipment, where the population proportion is 0.07, and the sample size is 200, what is the probability that the sample proportion will be at least 0.09?
(a) 0.3665 (b) 0.1333 (c) 0.8667 (d) 0.6335
Figured out this one, I think,

Z < (.07 - .09) / (.07 * .93 / 200) = -1.108

p = .1338

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