A survey of 200 college students showed the distribution of the number of hours of TV...

A survey of 200 college students showed the distribution of the number of hours of TV watched per week. The mean was found to be 3.66 hours with a standard deviation of 4.93. Assume the normal model is appropriate. Find the number of hours of TV watched by top 25% of college students.

Solutions

Expert Solution

Given that,

mean = = 3.66

standard deviation = = 4.93

n = 200

= 3.66

= / n =4.93 / 200=0.3486

Using standard normal table,

P(Z > z) = 25%

= 1 - P(Z < z) = 0. 25

= P(Z < z ) = 1 - 0.25

= P(Z < z ) = 0.75

= P(Z <0.67 ) = 0.

z = 0.67 using z table

Using z-score formula

= z * +

= 0.67 *0.3486+3.66

= 3.8936

= 4

orchestra answered 1 year ago

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