Hi, I am having a lot of trouble with Normal Distribution. I do online school and this lesson didn't really explain how to exactly complete the following problems. My teacher said that the answers to the problems below were all incorrect. I would greatly appreciate any help given!

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Consider this scenario for questions 5 - 8.

A standardized test was given to a set of high school juniors and the distribution of the data is bell shaped. The mean score is 800 and the standard deviation is 120.

5. Between which two scores did 95% of the students score?
z = -1.96
x = μ+σ*z
x = 74+8(-1.96)
x = 58.32
z = +1.96
x = μ+σ*z
x = 74+8(1.96)
x = 89.68
95% lie between 58 and 90

6. To qualify for a special summer camp for accelerated students, a student must score within the top 16% of all scores on the test. What score must a student make to qualify for summer camp?
z = +1.96
x = μ+σ*z
x = 74+8(1.96)
x = 89.68
90

7. What score is 1/2 standard deviation above the mean?
z = 1/2
score = 800 + (1/2)120 = 860

8. A student scores 900 on the test. How many more points did the student need to qualify for summer camp?
5/
(800 - 240 , 800+240)
6/
920
7/
800-240
560

Consider the following scenario for questions 9-12.

On the average, members at a local fitness center work out for 90 minutes with a standard deviation of 15 minutes. The distribution is normal.

9. What percentage of the fitness club members work out for 45 minutes or less?
z(45) = (45-90)/15 = -45/15 = -3
P(x <= 45) = P(z <= -3) = (-100,-3)
0.3%

10. What percentage of the fitness club members work out for 2 hours and 15 minutes or more?
z = (135 - 90)/15
1. 3%

11. 68% of the fitness club members work out between which two time intervals?
P(Z < z) = 0.84
0.5+0.34
68% within 1 sd
between 75 and 105 minutes

12. We can say that 99.7% of the fitness club members work out for no more than 135 minutes.
(Mean - 3 × std) and (Mean + 3 × std)
(90 - 3 × 15) and (90 + 3 × 15)
45 – 135

If you plug some of your numbers in at this web site, I think you'll soon see how things work.

http://davidmlane.com/hyperstat/z_table.html

Let's go through each problem and provide step-by-step solutions:

5. Between which two scores did 95% of the students score?
To find the scores between which 95% of students scored, we need to find the z-scores corresponding to the upper and lower boundaries and then convert them back to raw scores.
- The lower boundary corresponds to a z-score of -1.96 (from the standard normal table).
- The upper boundary corresponds to a z-score of 1.96 (from the standard normal table).
To convert these z-scores back to raw scores:
Lower score: x = μ + σ * z = 800 + (120 * -1.96) = 572.8 (rounded to 573)
Upper score: x = μ + σ * z = 800 + (120 * 1.96) = 1027.2 (rounded to 1027)
Therefore, 95% of the students scored between 573 and 1027.

6. To qualify for a special summer camp for accelerated students, a student must score within the top 16% of all scores on the test. What score must a student make to qualify for summer camp?
To find the score that corresponds to the top 16%, we need to find the z-score that corresponds to this percentile and then convert it back to a raw score.
- The top 16% corresponds to a percentile of 100 - 16 = 84%.
- To find the z-score corresponding to the 84th percentile, we look up this value in the standard normal table and find z = 0.994.
To convert this z-score back to a raw score:
x = μ + σ * z = 800 + (120 * 0.994) = 920.28 (rounded to 920)
Therefore, a student must score 920 or above to qualify for summer camp.

7. What score is 1/2 standard deviation above the mean?
To find the score that is 1/2 standard deviation above the mean, we simply add (1/2) * standard deviation to the mean.
score = mean + (1/2) * standard deviation = 800 + (1/2) * 120 = 860
Therefore, the score that is 1/2 standard deviation above the mean is 860.

8. A student scores 900 on the test. How many more points did the student need to qualify for summer camp?
To find the difference between the student's score and the score required to qualify for summer camp, we subtract the required score (920) from the student's score (900).
Difference = 900 - 920 = -20
Therefore, the student needed 20 more points to qualify for summer camp.

9. What percentage of the fitness club members work out for 45 minutes or less?
To find the percentage of fitness club members who work out for 45 minutes or less, we need to find the z-score corresponding to 45 minutes and calculate the corresponding area under the normal curve.
z = (x - μ) / σ = (45 - 90) / 15 = -3
Using the standard normal table, we find that the area to the left of -3 is approximately 0.0013 (or 0.13%).
Therefore, approximately 0.13% of fitness club members work out for 45 minutes or less.

10. What percentage of the fitness club members work out for 2 hours and 15 minutes or more?
To find the percentage of fitness club members who work out for 2 hours and 15 minutes (or 135 minutes) or more, we need to find the z-score corresponding to 135 minutes and calculate the area under the normal curve to the right of this z-score.
z = (x - μ) / σ = (135 - 90) / 15 = 3
Using the standard normal table, we find that the area to the right of 3 is approximately 0.0013 (or 0.13%).
Therefore, approximately 0.13% of fitness club members work out for 2 hours and 15 minutes or more.

11. 68% of the fitness club members work out between which two time intervals?
To find the time intervals between which 68% of fitness club members work out, we need to consider the area under the normal curve within 1 standard deviation from the mean on both sides.
Since the normal distribution is symmetric, 34% of the members work out below this range, while another 34% work out above this range.
Therefore, 68% of the members work out between the mean minus 1 standard deviation (90 - 15 = 75) and the mean plus 1 standard deviation (90 + 15 = 105) minutes.
So, 68% of fitness club members work out between 75 and 105 minutes.

12. We can say that 99.7% of the fitness club members work out for no more than 135 minutes.
To find the range within which 99.7% of fitness club members work out, we need to consider 3 standard deviations from the mean on both sides.
The range is given by (mean - 3 * standard deviation) to (mean + 3 * standard deviation).
Range = (90 - 3 * 15) to (90 + 3 * 15) = 45 to 135
Therefore, we can say that 99.7% of fitness club members work out for no more than 135 minutes.

Hi! I'll be happy to help you with your Normal Distribution problems.

For question 5, you're given a bell-shaped distribution with a mean score of 800 and a standard deviation of 120. You want to find the range within which 95% of the students scored. To solve this, you need to use the formula for z-score and the corresponding z-values for the desired percentage. In this case, 95% lies between the z-scores of -1.96 and 1.96.

To find the scores, you can use the formula x = μ + σ * z, where x is the score, μ is the mean, σ is the standard deviation, and z is the z-score.

For the lower score, x = 800 + 120 * (-1.96) = 58.32 (approximated to 2 decimal places).
For the upper score, x = 800 + 120 * 1.96 = 89.68 (approximated to 2 decimal places).

Therefore, 95% of the students scored between 58 and 90.

For question 6, you're asked to find the score necessary to qualify for a special summer camp, which is in the top 16% of all scores. To do this, you need to find the z-score corresponding to the given percentage (16%), and then use the formula x = μ + σ * z.

The z-score can be found using a z-table or a calculator, and for a percentage of 16%, the z-score is approximately 1.96 (positive because it represents the upper percentage).

Substituting the values into the formula, x = 800 + 120 * 1.96 = 89.68 (approximated to 2 decimal places).

Therefore, a student must score at least 90 to qualify for the summer camp.

For question 7, you're asked to find the score that is 1/2 standard deviation above the mean. Since the mean is 800 and the standard deviation is 120, you can calculate the score using the formula: score = mean + (1/2) * standard deviation.

Plugging in the values, score = 800 + (1/2) * 120 = 860.

Therefore, the score that is 1/2 standard deviation above the mean is 860.

For question 8, you're given a student's score of 900 and asked to find how many more points the student needed to qualify for the summer camp. To do this, you need to calculate the range within which the student needed to score to qualify for the summer camp.

Using the formula: range = (mean - 3 * standard deviation) and (mean + 3 * standard deviation), you can calculate the range as (800 - 3 * 120) and (800 + 3 * 120) = (440, 1160).

Since the student scored 900, the difference between the score and the upper end of the range is 1160 - 900 = 260.

Therefore, the student needed 260 more points to qualify for the summer camp.

For questions 9-12, you're given a different scenario with the average workout time and standard deviation at a fitness center. Your approach is correct and I don't see any need for additional explanation. Your answers for questions 9-12 are correct.

I hope this helps you with your Normal Distribution problems! If you have any further questions, feel free to ask!