The third grade class collected coins for the week.Class one raised 34.00. Class two raised 45.00 .Class three raised 76.00.How
2 answers:
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Answer:
a) 22497.7 < μ< 24502.3
b) With 99% confidence the possible error will not exceed 1002.3
Step-by-step explanation:
Given that:
Mean (μ) = 23500 kilometers per year
Standard deviation (σ) = 3900 kilometers
Confidence level (c) = 99% = 0.99
number of samples (n) = 100
a) α = 1 - c = 1 - 0.99 = 0.01
Using normal distribution table, is the z value of 1 - 0.005 = 0.995 of the area to the right which is 2.57.
The margin of error (e) is given as:
The 99% confidence interval = (μ - e, μ + e) = (23500 - 1002.3, 23500 + 1002.3) = (22497.7, 24502.3)
Confidence interval = 22497.7 < μ< 24502.3
b) With 99% confidence the possible error will not exceed 1002.3