$1,200 is received at the beginning of year 1, $2,200 is received at the beginning of year 2, and $3,300 is received at the begi
1 answer:
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Answer:
1) We need a random sample. For this case we assume that the sample selected was obtained using the simple random sampling method.
2) We need to satisfy the following inequalities:
So then we satisfy this condition
3) 10% condition. For this case we assume that the random sample selected n represent less than 10% of the population size N . And for this case we can assume this condition.
So then since all the conditions are satisfied we can conclude that we can apply the normal approximation given by:
So then the answer for this case would be:
a. Yes.
Explanation:
For this case we assume that the question is: If in the experiment described we can use the normal approximation for the proportion of interest.
For this case we have a sample of n =25
And we are interested in the proportion of registered students that spend more than 20 minutes to get to school.
X = 13 represent the number of students in the sample selected that have a time more than 20 min.
And then the estimated proportion of interest would be:
And we want to check if we can use the normal approximation given by:
So in order to do this approximation we need to satisfy some conditions listed below:
1) We need a random sample. For this case we assume that the sample selected was obtained using the simple random sampling method.
2) We need to satisfy the following inequalities:
So then we satisfy this condition:
3) 10% condition. For this case we assume that the random sample selected n represent less than 10% of the population size N . And for this case we can assume this condition.
So then since all the conditions are satisfied we can conclude that we can apply the normal approximation given by:
So then the answer for this case would be:
a. Yes.
Answer:
(a)Let X1 be the number of economy tires and X2 be the number of premium tires.
Objective function:
Maximize Z, where Z = 12X1 + 10X2
Subject to constraints
4X1/3 + X2/2 <= 600
4X1/5 + X2 <= 650
X1/2 + 2X/4 <= 580
X1/5 + X2/3 <= 120
X1, X2 = Z
(b) Check attachment for spreadsheet
(c) The maximum profit that can be obtained is $6032
