Download Home Work 4 Solutions

Home Work 4 Answer Key
1. The weights of the eggs produced by a certain breed of hen are normally distributed with mean 65
grams and standard deviation of 5 grams.
(a) What is the probability that one egg selected at random from a hen house will weigh more than 68
grams?
Normcdf(68, 200, 65, 5) = 0.2743
(b) What is the probability that one egg selected at random from a hen house will weigh less than 63
grams?
Normcdf(0, 63, 65, 5) = 0.3446
(c) What is the probability that one egg selected at random from a hen house will weigh between 64
and 67 grams?
Normcdf(64, 67, 65, 5) = 0.2347
(d) Consider a carton of 12 eggs to be a simple random sample (SRS) of hen's eggs. What is the
probability that the average weight of the 12 eggs in a carton selected at random will be more than 68
grams?
Normcdf(68, 200, 65, 5/sqrt(12)) = 0.0186
2. Carbon monoxide (CO) emissions for a certain kind of car vary according to a normal distribution
with mean 2.9 g/mi and a standard deviation of 0.4. A car dealership has 60 of these cars.
a) What is the probability that the average carbon monoxide emissions of these 60 cars is between
3.0 and 3.1 g/mi?
Normcdf(3, 3.1, 2.9, 0.4/sqrt(60)) = 0.0272
b) A customer wants to buy one of these cars under the condition that the car’s carbon monoxide
emission is not in the top 25% of carbon monoxide emissions of such cars. What is the highest level
of carbon monoxide emission that is acceptable for the customer?
INVNORM(0.75, 2.9, 0.4) =3.17
3. Because of slight irregularities that constantly occur on a production line, cars of identical make
are not identical in every fashion. One way in which they vary is their gas mileage. Consider the new
Toyota RV4. The car's mechanical operation is based on a current Toyota model; engineers expect
the variation in gas mileage from car to car to have a normal distribution with standard deviation
0.35. What they don't know is the average gas mileage for the RV4. To find out, they take an SRS of
32 RV4s and drive each, recording gas mileage (mpg). The sample mean is 35.28.
a) Use this data to construct a 94% CI for the mean gas mileage of all RV4s.
35.28-(1.881*(0.35)/sqrt(32)) = 35.164;
35.28+(1.881*(0.35)/sqrt(32)) = 35.396
b) Compute a 98% confidence interval (same data).
35.28-(2.327*(0.35)/sqrt(32)) = 35.136;
35.28+(2.327*(0.35)/sqrt(32)) = 35.424
c) How do the margins of error for your two intervals compare?
The 98% confidence interval has a bigger margin of error (wider interval)
4. The cholesterol levels of a random sample of 120 men are measured. The sample mean is 188 and
the sample standard deviation is 40. Find the following:
a) A 92% confidence interval for the true mean of cholesterol level of men
188-(1.751*(40)/sqrt(120)) = 181.606;
188+(1.751*(40)/sqrt(120)) = 194.393
b) The margin of error
margin of error = 6.394
c) State three ways that help you reduce the margin of error
Lower the confidence level
Increase the sample size
Decrease the standard deviation (apply appropriate sampling method, address outliers, etc)