Distribution 67074

The time required to complete the test has a normal distribution with a mean of 50 minutes and a standard deviation of 10 minutes. What percentage of students take the test within 30 minutes?

Correct answer:

p =  2.5 %

Step-by-step explanation:

m=50 min σ=10 min t=30 min  t = m  2 σ  p=5095/2=2.5%

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