** 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

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.

a. To find the percentage of students who texted for fewer than 50 minutes, we can use the standardized normal distribution.

Step 1: Standardize the value using the formula z = (x - μ) / σ, where x is the value (50 minutes in this case), μ is the mean (60 minutes), and σ is the standard deviation (20 minutes).

z = (50 - 60) / 20
z = -0.5

Step 2: Find the corresponding area under the standard normal curve using a z-table or a calculator. The area to the left of -0.5 represents the percentage of students who texted for fewer than 50 minutes.

Look up the value of -0.5 in the z-table, and you will find that the area to the left is approximately 0.3085.

Therefore, approximately 30.85% of students texted for fewer than 50 minutes.

b. To find the probability of selecting at random a student who spends between 10 and 30 minutes texting, we need to find the area under the normal curve between the two values.

Step 1: Standardize the values.

z1 = (10 - 60) / 20
z1 = -2.5

z2 = (30 - 60) / 20
z2 = -1.5

Step 2: Find the difference between the areas to the left of z2 and z1 using a z-table or a calculator.

Lookup the value of -1.5 in the z-table, and you will find that the area to the left is approximately 0.0668.
Lookup the value of -2.5 in the z-table, and you will find that the area to the left is approximately 0.0062.

The difference between these two values represents the probability that a student spends between 10 and 30 minutes texting.

0.0668 - 0.0062 = 0.0606

Therefore, the probability of selecting at random a student who spends between 10 and 30 minutes texting is approximately 0.0606.

c. To find 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, we need to find the cumulative probability of both ends.

Step 1: Find the probability of spending less than 10 minutes.

Standardize the value:
z1 = (10 - 60) / 20
z1 = -2.5

Lookup the value of -2.5 in the z-table, and you will find that the area to the left is approximately 0.0062.

Step 2: Find the probability of spending more than 110 minutes.

Standardize the value:
z2 = (110 - 60) / 20
z2 = 2.5

Lookup the value of 2.5 in the z-table, and you will find that the area to the left is approximately 0.9938.

The sum of these two areas represents the probability of selecting a student who spends an extreme amount of time texting.

0.0062 + (1 - 0.9938) = 0.0124

Therefore, the probability of selecting at random a student who spends an extreme amount of time texting is approximately 0.0124.

Let's continue with the remaining questions in a separate response.