a. Either-or probabilities are found by adding the individual probabilities.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability. Add the two.
b. If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
Use same equation and table, but this time multiply.
c. Same process as B.
d. Same as A, but looking for % below.
e. 95% = mean ± Z (SD)
Use table to find Z for ±.025.
Does anyone know how to solve for this
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 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. -
b. the probability of selecting at random (with replacement) two students who spent a below-average amount of time texting.
c. the probability of selecting at random (with replacement) two students who spent more than 75 minutes texting.
d. the percentile rank of a student who spent 100 minutes texting.
e. the two numbers of minutes that define the middle 95% of students in the distribution.
1 answer
1 answer