Question

1. Sleep time per night among all university students is approximately normally distributed with a mean...

1. Sleep time per night among all university students is approximately normally distributed with a mean of 6.78 hours and a standard deviation of 1.24 hours. A random sample of 31 students is selected.

Question: Calculate the value that 20 percent of the means would exceed.

  

Math   Statistics-And-Probability   
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Homework Answers

Answer #1

Solution :

Given that,

mean = = 6.78

standard deviation = = 1.24

n = 31

= 6.78

= / n = 1.24 / 31 = 0.2227

P(Z > z) = 0.20

1 - P(Z < z) = 0.20

P(Z < z) = 1 - 0.20

P(Z < 0.84) = 0.80

z = 0.84

= z * + = 0.84 * 0.2227 + 6.78 = 6.97

Value = 6.97

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