Answer:
Option C) 70.5 and 73.5 minutes
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 72 minutes
Standard Deviation, σ = 4 minutes
Sample size, n = 64
We are given that the distribution of amount of time is a bell shaped distribution that is a normal distribution.
Empirical Formula:
Standard error due to sampling =
[tex]S.E = \dfrac{\sigma}{\sqrt{n}} = \dfrac{4}{\sqrt{64}} = 0.5[/tex]
Thus, by empirical formula 99.7% of data will lie between:
[tex]\mu - 3(S.E) = 72-3(0.5) = 70.5\\\mu + 3(S.E) = 72+3(0.5) = 73.5[/tex]
There would be a 99.7% chance that the sample mean amount of time taken on the exam would be between 70.5 and 73.5 minutes.
Thus, the correct answer is
Option C) 70.5 and 73.5 minutes
99.7% of time taken on the exam would be between 60 and 84 minutes
The three sigma rule states that for a normal distribution, 68% are within one standard deviation from the mean, 95% are within two standard deviation from the mean and 99.7% are within three standard deviation from the mean.
Therefore given that mean (μ) = 72, standard deviation (σ) = 4;
99.7% are within three standard deviation from the mean = μ ± 3σ = 72 ± 3*4 = (60, 84)
99.7% of time taken on the exam would be between 60 and 84 minutes
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