Question

1. Assume the standard deviation is 1.5 hours—
2. What is the probability of a random person in your age group getting less than 6 hours of sleep?
3. What is the probability of a random person in your age group getting between 7 - 9 hours of sleep?
4. What is the probability of a random person in your age group getting more than 9 hours of sleep?
5. Are any of the probabilities in questions 5 - 7 considered unusual? Explain.

age group amount of sleep is = 8hrs

Answer 1

Solution :

Given that ,

mean = = 8 hours

standard deviation = = 1.5 hours

5) P(x < 6) = P[(x - ) / < (6 - 8) / 1.5]

= P(z < -1.33)

Using z table,

= 0.0918

6) P( 7 < x < 9 ) = P[(7 - 8)/ 1.5 ) < (x - ) /  < (9 - 8) / 1.5) ]

= P(-0.67 < z < 0.67)

= P(z < 0.67) - P(z < -0.67)

Using z table,

= 0.7486 - 0.2514

= 0.4972

7) P(x > 9) = 1 - p( x< 9)

=1- p P[(x - ) / < (9 - 8) / 1.5 ]

=1- P(z < 0.67)

Using z table,

= 1 - 0.7486

= 0.2514

No, Are any of the probabilities in questions 5 - 7 not considered unusual, because all probabilities more than 5%