Asked by Crystal
** I already finish these questions
Can you answer so I can compare my final answers w/ yours.**
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.
a. the percentage of students who texted for fewer than 50 minutes.
b. the probability of selecting at random a student who spends between 10 and 30 minutes texting.
c. 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.
d. the probability of selecting at random (with replacement) two students who spent a below-average amount of time texting.
e. the probability of selecting at random (with replacement) two students who spent more than 75 minutes texting.
f. the percentile rank of a student who spent 100 minutes texting.
the two numbers of minutes that define the middle 95% of students in the distribution
Can you answer so I can compare my final answers w/ yours.**
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.
a. the percentage of students who texted for fewer than 50 minutes.
b. the probability of selecting at random a student who spends between 10 and 30 minutes texting.
c. 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.
d. the probability of selecting at random (with replacement) two students who spent a below-average amount of time texting.
e. the probability of selecting at random (with replacement) two students who spent more than 75 minutes texting.
f. the percentile rank of a student who spent 100 minutes texting.
the two numbers of minutes that define the middle 95% of students in the distribution
Answers
Answered by
John
50-60/20 = -.5 z score Look that up to find the % You want z<-.5
10-60/20 30-60/20
-2.5 and -1.5 are the z-scores find the % by finding
-2.5<z<-1.5
Can you submit your answers so they can be checked.
10-60/20 30-60/20
-2.5 and -1.5 are the z-scores find the % by finding
-2.5<z<-1.5
Can you submit your answers so they can be checked.
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