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.
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.
To solve these probability questions, we will use the normal distribution and z-scores. The z-score measures how many standard deviations an individual data point is from the mean. We can then use a standard normal distribution table or a calculator to find the probabilities.
a. 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 need to find the area under the normal curve outside the range of (10, 110).
First, we need to convert these values into z-scores using the formula:
z = (x - mean) / standard deviation
For less than 10 minutes, the z-score is:
z1 = (10 - 60) / 20 = -2.5
For more than 110 minutes, the z-score is:
z2 = (110 - 60) / 20 = 2.5
Next, using a standard normal distribution table or a calculator, we can find the area under the curve corresponding to these z-scores. The area to the left of -2.5 is the same as the area to the right of 2.5.
P(z < -2.5) + P(z > 2.5) = P(z > 2.5) [because the area is symmetric]
Using a standard normal distribution table or a calculator, you can find the probability corresponding to the z-score of 2.5 to get the answer.
b. 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 one student below average and then multiply by itself since we are selecting two students.
The z-score for below-average is:
z = (x - mean) / standard deviation
Substituting in the values:
z = (60 - 60) / 20 = 0
Using the standard normal distribution table or a calculator, you can find the probability corresponding to a z-score of 0. Then, multiply this probability by itself to find the probability of selecting two students below average.
c. To find the probability of selecting two students who spent more than 75 minutes texting, we need to find the probability of selecting one student above 75 and then multiply by itself since we are selecting two students.
The z-score for more than 75 minutes is:
z = (x - mean) / standard deviation
Substituting in the values:
z = (75 - 60) / 20 = 0.75
Using the standard normal distribution table or a calculator, you can find the probability corresponding to a z-score of 0.75. Then, multiply this probability by itself to find the probability of selecting two students who spent more than 75 minutes.
d. To find the percentile rank of a student who spent 100 minutes texting, we need to find the area under the normal curve to the left of the data point.
The z-score can be calculated as:
z = (x - mean) / standard deviation
Substituting in the values:
z = (100 - 60) / 20 = 2
Using the standard normal distribution table or a calculator, you can find the probability corresponding to a z-score of 2. This probability represents the percentile rank of the student who spent 100 minutes texting.
e. To find the two numbers of minutes that define the middle 95% of students in the distribution, we need to find the z-scores corresponding to the upper and lower ends of the 95% range.
Since we want to find the middle 95%, we use a z-score that leaves 2.5% on both tails. The corresponding z-scores can be found using the standard normal distribution table or a calculator.
The z-score for the lower end is:
z1 = -1.96
The z-score for the upper end is:
z2 = 1.96
To convert these z-scores back into the original distribution, we can use the formula:
x = (z * standard deviation) + mean
Substituting in the values:
x1 = (-1.96 * 20) + 60
x2 = (1.96 * 20) + 60
The two numbers of minutes that define the middle 95% of students in the distribution are x1 and x2.
Remember to use a standard normal distribution table or a calculator to find the probabilities corresponding to the z-scores.