Can you please tell me how to solve for the following?

Pedro took an exam in a class in which the mean was 64 with a standard deviation of 6. If his z score was +3, what was his exam score?

A students commute to school is normally distributed with a mean of 31 min and a standard deviation of 7 min. What is the probability that the student gets to school in 17 to 31 min?

Are the formulas the same for these?

And last one I need help with this:

Assume a normally distributed set of test scores with a mean of 100 and a standard deviation of 20. Find the probability that a person selected at random will have a score of the following:

a-between 60 and 140
b-less than 60

THANK YOU! I just need to know how to solve please.

Equations are the same.

Z = (score-mean)/SD

For the second one, find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z scores.

For the last problem, use the same equation and table.

Sure! I will guide you through the steps to solve each of these questions.

1. Pedro's Exam Score:
To find Pedro's exam score, we can use the formula for converting a z-score to a raw score. The formula is:
Raw Score = (Z-score * Standard Deviation) + Mean

Given that Pedro's z-score is +3, the mean is 64, and the standard deviation is 6, we can substitute these values into the formula:
Raw Score = (3 * 6) + 64
Raw Score = 18 + 64
Raw Score = 82

Therefore, Pedro's exam score is 82.

2. Probability of Commute Time:
To find the probability that the student gets to school in 17 to 31 minutes, we need to calculate the standard deviation of the z-scores. The formula for converting raw scores to z-scores is:
Z-score = (X - Mean) / Standard Deviation

First, we calculate the z-score for 31 minutes:
Z-score_31 = (31 - 31) / 7
Z-score_31 = 0

Next, we calculate the z-score for 17 minutes:
Z-score_17 = (17 - 31) / 7
Z-score_17 = -2

Once we have the z-scores, we can use a z-table or statistical software to find the probabilities associated with these z-scores. The probability of the student getting to school in 17 to 31 minutes is the area under the normal distribution curve between these two z-scores.

3. Probability of Test Scores:
The formula for finding probabilities associated with test scores is similar. We can use z-scores to find the probability. However, in this case, we are given the desired scores and we need to convert them into z-scores first.

a. Probability of score between 60 and 140:
To find the probability of a score between 60 and 140, we first need to convert these scores into z-scores using the formula:
Z-score = (X - Mean) / Standard Deviation

For a score of 60:
Z-score_60 = (60 - 100) / 20
Z-score_60 = -2

For a score of 140:
Z-score_140 = (140 - 100) / 20
Z-score_140 = 2

Once we have the z-scores, we can use a z-table or statistical software to find the probabilities associated with these z-scores. The probability of having a score between 60 and 140 is the area under the normal distribution curve between these two z-scores.

b. Probability of score less than 60:
To find the probability of a score less than 60, we convert the given score into a z-score using the same formula:
Z-score_60 = (60 - 100) / 20
Z-score_60 = -2

Again, we can use a z-table or statistical software to find the probability associated with this z-score. The probability of having a score less than 60 is the area under the normal distribution curve to the left of this z-score.

I hope this step-by-step explanation helps you to understand how to solve these problems. Let me know if you have any further questions!

Sure! I can help you with these questions.

Let's start with the first question:

Pedro took an exam in a class in which the mean was 64 with a standard deviation of 6. If his z score was +3, what was his exam score?

To solve for Pedro's exam score, you can use the formula for z-score:

z = (x - mean) / standard deviation

In this case, Pedro's z-score is given as 3, and the mean is 64, and the standard deviation is 6. Rearranging the formula, we get:

x = z * standard deviation + mean

Plugging in the values, we can calculate Pedro's exam score:

x = 3 * 6 + 64 = 18 + 64 = 82

So, Pedro's exam score was 82.

Moving on to the second question:

A student's commute to school is normally distributed with a mean of 31 min and a standard deviation of 7 min. What is the probability that the student gets to school in 17 to 31 min?

To find the probability in this range, we need to standardize the values and use the standard normal distribution table (also known as the z-table).

We can convert the lower and upper bounds of the range into z-scores using the formula:

z = (x - mean) / standard deviation

For the lower bound (17 min):

z1 = (17 - 31) / 7 = -14 / 7 = -2

For the upper bound (31 min):

z2 = (31 - 31) / 7 = 0

Now, we need to find the area under the standard normal curve between these two z-scores. We can do this by looking up the corresponding probabilities in the z-table.

The probability of z being less than or equal to -2 is 0.0228, and the probability of z being less than or equal to 0 is 0.5000.

To find the probability between -2 and 0, we subtract the probability at -2 from the probability at 0:

P(-2 < z < 0) = 0.5000 - 0.0228 = 0.4772

So, the probability that the student gets to school in 17 to 31 min is 0.4772.

For your third question:

Are the formulas the same for these?

Yes, the formulas used to solve these questions are similar. In both cases, we use the concept of z-scores and the formula z = (x - mean) / standard deviation. However, the specific calculations may differ depending on the given values and the question being asked.

Now, let's move on to the last question:

Assume a normally distributed set of test scores with a mean of 100 and a standard deviation of 20. Find the probability that a person selected at random will have a score of the following:

a) Between 60 and 140
b) Less than 60

To find these probabilities, we can again use z-scores and the standard normal distribution table.

a) Between 60 and 140:

First, find the z-scores for the lower and upper bounds:

For the lower bound (60):

z1 = (60 - 100) / 20 = -40 / 20 = -2

For the upper bound (140):

z2 = (140 - 100) / 20 = 40 / 20 = 2

Using the z-table, find the probabilities for these z-scores:

P(-2 < z < 2) = P(z < 2) - P(z < -2)

The probability of z being less than 2 is 0.9772, and the probability of z being less than -2 is 0.0228.

P(-2 < z < 2) = 0.9772 - 0.0228 = 0.9544

So, the probability that a person selected at random will have a score between 60 and 140 is 0.9544.

b) Less than 60:

To find the probability of a score less than 60, we need to find the probability of z being less than the z-score corresponding to 60.

z = (60 - 100) / 20 = -40 / 20 = -2

The probability of z being less than -2 is 0.0228.

So, the probability that a person selected at random will have a score less than 60 is 0.0228.

I hope this helps! Let me know if you have any further questions.