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Answer:
a) Margin of error = 166.311
b) sample size ≥ 62
Step-by-step explanation:
Given:
Average cost = $1599
Sample size = 50 persons
Standard deviation = $600
Confidence level is 95%
a) Margin of error =
Now for confidence level of 95% z-value = 1.96
Thus,
Margin of error =
or
Margin of error = 166.311
b) For Margin of error ≤ 150
≤ 150
or
≤ 150
or
≤ √n
or
√n ≥ 7.84
or
n ≥ 61.4656
Therefore,
sample size ≥ 62
Answer:
7,200
Step-by-step explanation:
Let's start by calculating how many students the university accepts. The problem says they accept 60% of the students. 60% as a decimal is 0.6. To calculate 60% of something, you multiply .60 by whatever quantity. In this case, it would be the students.
Number of accepted students = = 18,000
The university accepts 18,000 students.
Now, the next problem is that of those accepted students, we need to find out how many actually enroll. The problem says they accept 40% of the accepted students. Applying the same principle,
Number of enrolled students = = 7,200
The number of students that actually enroll is 7,200.
Alternatively, would give you the same result.
Combining the pieces together, we are asking for 60% of 30,000. Then getting 40% of that 60%.