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
Z = (score-mean)/SD
Use table In back of text labeled area under normal distribution.
C. Either-or probabilities found by adding individual probabilities.
D,E. Both probabilities found by multiplying individual probabilities.
To solve these questions, we will use the concept of the standard normal distribution. A standard normal distribution has a mean of 0 and a standard deviation of 1. We can convert any value from a normal distribution with mean μ and standard deviation σ to a standard normal distribution using the formula: Z = (X - μ) / σ, where Z is the z-score of the value X.
a. To find the percentage of students who texted for fewer than 50 minutes, we need to calculate the z-score of 50 using the formula Z = (X - μ) / σ. Here, X is 50, μ is 60, and σ is 20.
Z = (50 - 60) / 20 = -0.5
Using a standard normal distribution table or a calculator, we can find the area to the left of Z = -0.5, which represents the percentage of students who texted for fewer than 50 minutes.
b. To find the probability of selecting a student who spends between 10 and 30 minutes texting, we need to calculate the z-score for each boundary value (10 and 30) using the formula Z = (X - μ) / σ.
For X = 10:
Z = (10 - 60) / 20 = -2.5
For X = 30:
Z = (30 - 60) / 20 = -1.5
Using the standard normal distribution table or a calculator, find the area to the left of Z = -1.5 and subtract the area to the left of Z = -2.5. This will give you the probability of selecting a student who spends between 10 and 30 minutes texting.
c. To find the probability of selecting a student who spends an extreme amount of time texting (less than 10 minutes OR more than 110 minutes), we need to calculate the z-scores for each boundary value (10 and 110) using the formula Z = (X - μ) / σ.
For X = 10:
Z = (10 - 60) / 20 = -2.5
For X = 110:
Z = (110 - 60) / 20 = 2.5
The probability of selecting a student who spends an extreme amount of time texting is the sum of the areas to the left of Z = -2.5 and to the right of Z = 2.5. You can find these probabilities using a standard normal distribution table or a calculator.
d. To find the probability of selecting two students who spent a below-average amount of time texting, we need to find the probability of selecting a student with a z-score less than 0 (below-average) and then multiply it by itself, since we are selecting two students (with replacement).
The probability of selecting a student with a z-score less than 0 can be found using a standard normal distribution table or a calculator. Suppose this probability is P.
The probability of selecting two students who spent a below-average amount of time texting is P * P.
e. To find the probability of selecting two students who spent more than 75 minutes texting, we need to find the probability of selecting a student with a z-score greater than the z-score for 75 (X = 75, μ = 60, σ = 20).
Z = (75 - 60) / 20 = 0.75
The probability of selecting a student with a z-score greater than 0.75 can be found using a standard normal distribution table or a calculator. Suppose this probability is Q.
The probability of selecting two students who spent more than 75 minutes texting is Q * Q.
f. To find the percentile rank of a student who spent 100 minutes texting, we need to find the area to the left of the z-score for 100 (X = 100, μ = 60, σ = 20).
Z = (100 - 60) / 20 = 2
Using the standard normal distribution table or a calculator, find the area to the left of Z = 2. This represents the percentile rank of the student who spent 100 minutes texting.
To find the two numbers of minutes that define the middle 95% of students in the distribution, we need to find the z-scores for the boundaries of this range.
Since the middle 95% corresponds to a range that covers 95% of the distribution, we need to find the z-scores that leave 2.5% on each tail of the distribution.
Using a standard normal distribution table or a calculator, find the z-scores that leave 2.5% on each tail.
The lower boundary can be found as follows:
Z_lower = -Z_upper
The upper boundary can be found directly from the z-score.
Finally, convert the z-scores back to the original distribution using the formula X = μ + Z * σ, where X is the number of minutes spent texting.
I hope this explanation helps! Let me know if you have any further questions.