Let x denote the average number of hours per week that a student at Happy High School spends doing homework. Suppose the frequency distribution of x is normal (i. E., symmetrical and bell-shaped) with a mean of 3.4 hrs. And a standard deviation of 2.6 hrs. What percentage of the students at Happy High spend an average of less than 6 hours per week doing homework?

0.84134

Step-by-step explanation:

Given that:

Mean = 3.4 hours

Standard deviation (s) = 2.6 hours

Percentage of students that spend less than 6 hours doing homework

P(x < 6)

Using the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (6 - 3.4) / 2.6

Z = 2.6 / 2.6

Z = 1

p(Z < 1) = 0.84134 ( Z probability calculator)

The percentage of the students at Happy High spends an average of less than 6 hours per week doing homework is 0.84134

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