Question

Suppose that a principal of a local high school tracks the number of minutes his students spend texting on a given school day. He finds that the distribution of minutes spent texting is roughly normal with a mean of 60 and a standard deviation of 20. Use this information to answer the following questions.

1. Based on the statistics, what is the probability of selecting at random a student who spends an extreme amount of time texting – either less than 10 minutes OR more than 110 minutes?

2. Based on the statistics, what is the probability of selecting at random (with replacement) two students who spent a below-average amount of time texting?

3. Based on the statistics, what is the probability of selecting at random (with replacement) two students who spent more than 75 minutes texting?

4. Based on the statistics, what is the percentile rank of a student who spent 100 minutes texting?

5. Based on the statistics, find the two numbers of minutes that define the middle 95% of students in the distribution. What is the value for the lower number that you found?

6. Based on the statistics, find the two numbers of minutes that define the middle 95% of students in the distribution. What is the value for the higher number that you found?

Answer 1

P(z<Z) table :

x = minutes spent texting

1.

P(extreme) = 1 - P(10<=x<=110)

P(10<=x<=110) :

P(10<=x<=110) = 0.9876

P(extreme) = 1 - P(10<=x<=110) = 1 - 0.9876 = 0.0124

P(extreme) = 0.0124

2.

P(below average) = P(z<0) = 0.50 {from table}

P(2 students below average) = 0.50*0.50 = 0.25

3.

P(x>=75) :

probability of selecting at random (with replacement) two students who spent more than 75 minutes texting

= P(x>=75)^2

= 0.2266^2

= 0.0513

4.

x=100

z = (100-60)/20 = 40/20 = 2

P(x<100) = P(z<2) = 0.9772

percentile = 97.72%