SOLUTION: The time to complete a standardized exam in the BYU-Idaho Testing Center is approximately normal with a mean of 70 minutes and a standard deviation of 10 minutes. Using the 68-95-9

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Question 1196804: The time to complete a standardized exam in the BYU-Idaho Testing Center is approximately normal with a mean of 70 minutes and a standard deviation of 10 minutes. Using the 68-95-99.7% Rule, approximately what percentage of students will complete the exam in under fifty minutes? Give your answer as a percentage accurate to one decimal place.
Answer by math_tutor2020(3817) About Me  (Show Source):
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Answer: 2.5%

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Explanation:

Given info
mu = 70 and sigma = 10
this represents the mean and standard deviation respectively

The 68-95-99.7% Rule is also known as the Empirical Rule.

That rule lays out three basic properties
  • Approximately 68% of the normal distribution is within 1 standard deviation of the mean.
  • Approximately 95% of the normal distribution is within 2 standard deviations of the mean.
  • Approximately 99.7% of the normal distribution is within 3 standard deviations of the mean.
Compute the z score for x = 50
z = (x - mu)/sigma
z = (50 - 70)/10
z = -20/10
z = -2
We're exactly 2 standard deviations below the mean.

We have roughly 95% of the values between z = -2 and z = 2
I.e. P(-2 < z < 2) = 0.95 approximately

That leaves 100% - 95% = 5% of the area for the tails to split up
Each tail gets (5%)/2 = 2.5%

Therefore, about 2.5% of the normally distributed population will have a z score smaller than -2
P(z < -2) = 0.025 approximately
P(x < 50) = 0.025 approximately when mu = 70 and sigma = 10