| Statistics Assignment #3 |
Show the given information and formulae. Make diagrams wherever possible:
1/ The mean of a random sample of size n = 36 is used to estimate the mean of a normal
population with s = 18. With what probability can we assert that the error of this estimate
will be less than 9 if we use The Central Limit Theorem?
(click here for solution) (5)
2/ A random sample of size n = 3 is to be chosen from a finite population of size N = 80.
Find the probability of each possible sample.
(click here for solution) (2)
3/ Random samples of size n = 2 are drawn from the finite population which consists
of the numbers 1, 3, 5, and 7. If µ = 4 and s = 
a) list the 6 possible samples of size n = 2 that can be drawn without replacement
from the given population and calculate their means.
b) Construct the sampling distribution of the mean for these samples.
c) Calculate the mean and standard deviation of this probability distribution.
(click here for solution) (10)
4/ What is the value of the finite population correction factor when:
| a) n = 30 and N = 120; | b) n = 20 and N = 500 |
(click here for solution) (2)
5/ When we sample from an infinite population, what happens to the
standard error of the mean if the sample size is:
| a) increased from 20 to 500 | b) decreased from 490 to 40 |
(click here for solution) (4)
6/ The time it takes a group of real estate salespersons to meet their annual quotas
is 10.3 months on average with a standard deviation of 1.6 months.
Find the probabilities that in a sample of 16 such salespersons, the average time
it will take them to meet their annual quotas will be:
| a) less than 10 months | b) more than 11.5 months |
(click here for solution) (4)
7/ How many different samples of size n = 4 can be selected from a finite population size:
| a) N = 18 | b) N = 30 | c) N = 100 ? |
(click here for solution) (3)
8/ The mean of a random sample of size n = 81 is used to estimate the mean annual
growth of certain plants. If s = 3.6 mm for such data, use the Central Limit Theorem
to find the probabilities that this estimate will be off either way by:
| a) less than 1.0 mm | b) less than 0.5 mm |
(click here for solution) (4)
9/ A random sample of 40 mechanics took an average of 24.05 min with
a standard deviation of 2.68 min to perform a certain task. Construct a 95%
confidence interval for the average time it takes such a mechanic to perform the task.
(click here for solution) (5)
10/ A random sample of 120 passengers arriving at Dorval airport took an
average of 24.15 minutes with s = 3.29 minutes to retrieve their luggage and get through customs.
a) construct a 95% confidence interval for the average time it takes to retrieve
luggage and get through customs.
b) By how much is the maximum error increased if we use a 99% level of confidence?
(click here for solution) (4)
11/ A random sample of 40 cans of pineapple slices has an estimated
mean weight of 15.85 ounces and a standard deviation of 0.23 ounces.
With what confidence can we assert that this estimate is "off" by at most 0.06 ounces?
(click here for solution) (4)
12/ A sample of 200 families showed that they spent an average of $218.67 per week
on food with a standard deviation of $14.93.
Construct a 90% confidence interval for this sample.
(click here for solution) (3)
TOTAL (50)
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Solutions
1) 
If z = 3, P(E < 9) = 2(.4987) = .9974
2) n = 3, N = 80 therefore P(each sample) = 1/ 80 C 3 = 1/ 82,160
| Sample | Sample Mean | Probability |
| (1, 3) | 2 | 1/6 |
| (1, 5) | 3 | 1/6 |
| (1, 7) | 4 | 2/6 |
| (3, 5) | 4 | |
| (3, 7) | 5 | 1/6 |
| (5, 7) | 6 | 1/6 |
4) a) 
5) a) 
b) 
6) a) s = 1.6/4 = 0.4, so z = (10 - 10.3)/0.4 = -0.75
since we want P(x < 10) we want the lower tail probability.
z = 0.75 tells us that .2734 is the belly probability, so the tail is 0.5 - .2734 = .2266
b) z = (11.5 - 10.3)/0.4 = 3
since we want P(x > 11.5) we want the upper tail probability = 0.5 - .4987 = .0013
| a) 18 C 4 = 3,060 samples | b) 30 C 4 = 27,405 samples | c) 100 C 4 = 3,921,225 samples |
8) a) S x = 3.6/9 = 0.4, so z = 1/0.4 = 2.5, and P(-2.5 < z < 2.5) = 2(.4938) = .9876
b) P(E < 0.5) = P(-1.25 < z < 1.25) = 2(.3944) = .7888
9) A 95% confidence interval puts 47.50% of the population on either side of the mean,
so z = ± 1.96.
Therefore, our confidence interval boundaries will be 1.96(S x) below and above the mean.
the confidence interval about µ is: 23.22 < µ < 24.88.
10) a) 95% confidence interval makes z = ± 1.96
the confidence interval about µ is: 23.56 < µ < 24.74
b) 99% confidence interval makes z = ± 2.575
0.77 - 0.59 = 0.18 minute increase in the maximum error.
11) 
. Confidence level is 2(.4505) = 90.1% confidence
12) 90% confidence interval makes z = ± 1.645
the confidence interval about µ is: 216.93 < µ < 220.41
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