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 labeled area under normal distribution in the back of your text.
Either-or probabilities found by adding individual probabilities.
Both probabilities found by multiplying individual probabilities.
To solve these problems, we will use the z-score formula and the standard normal distribution table. The z-score formula is given by:
z = (x - μ) / σ
where x is the value, μ is the mean, and σ is the standard deviation.
a. To find the percentage of students who texted for fewer than 50 minutes, we calculate the z-score:
z = (50 - 60) / 20
z = -0.5
Using the standard normal distribution table, we find the area to the left of z = -0.5 is approximately 0.3085. Therefore, the percentage of students who texted for fewer than 50 minutes is 30.85%.
b. To find the probability of selecting a student who spends between 10 and 30 minutes texting, we calculate the z-scores for both values:
z1 = (10 - 60) / 20
z1 = -2.5
z2 = (30 - 60) / 20
z2 = -1.5
Using the standard normal distribution table, we find the area to the left of z = -2.5 is approximately 0.0062 and the area to the left of z = -1.5 is approximately 0.0668. Therefore, the probability of selecting a student who spends between 10 and 30 minutes texting is 0.0668 - 0.0062 = 0.0606.
c. To find the probability of selecting a student who spends an extreme amount of time texting, either less than 10 minutes OR more than 110 minutes, we calculate the z-scores:
z1 = (10 - 60) / 20
z1 = -2.5
z2 = (110 - 60) / 20
z2 = 2.5
Using the standard normal distribution table, we find the area to the left of z = -2.5 is approximately 0.0062 and the area to the left of z = 2.5 is approximately 0.9938. Therefore, the probability of selecting a student who spends an extreme amount of time texting is 0.0062 + (1 - 0.9938) = 0.0124.
d. To find the probability of selecting two students who spent a below-average amount of time texting, we need to find the below-average probability for one student, and then square it for two students. The below-average z-score is:
z = (60 - 60) / 20
z = 0
Using the standard normal distribution table, we find the area to the left of z = 0 is 0.5000. Therefore, the probability of selecting a student who spent a below-average amount of time texting is 0.5000. To find the probability for two students, we square this value: 0.5000 * 0.5000 = 0.2500.
e. To find the probability of selecting two students who spent more than 75 minutes texting, we calculate the above-average probability for one student, and then square it for two students. The above-average z-score is:
z = (75 - 60) / 20
z = 0.75
Using the standard normal distribution table, we find the area to the left of z = 0.75 is approximately 0.7734. Therefore, the probability of selecting a student who spent more than 75 minutes texting is 1 - 0.7734 = 0.2266. To find the probability for two students, we square this value: 0.2266 * 0.2266 = 0.0514.
f. To find the percentile rank of a student who spent 100 minutes texting, we calculate the z-score:
z = (100 - 60) / 20
z = 2
Using the standard normal distribution table, we find the area to the left of z = 2 is approximately 0.9772. Therefore, the percentile rank of a student who spent 100 minutes texting is 97.72%.
The two numbers of minutes that define the middle 95% of students in the distribution can be calculated by finding the z-scores at the cutoff values. Since the distribution is symmetric, we can calculate for the upper and lower cutoff values separately:
Lower cutoff value:
z1 = (1 - 0.95) / 2 = -0.025
Upper cutoff value:
z2 = 0.95 + (-0.025) = 0.925
Solving for the corresponding values of z using the standard normal distribution table:
Lower cutoff value:
x1 = (z1 * 20) + 60
Upper cutoff value:
x2 = (z2 * 20) + 60
Therefore, the two numbers of minutes that define the middle 95% of students in the distribution are x1 and x2.