(10 pts) Using diaries for many weeks, a study on the lifestyles of visually impaired students was conducted. The students kept track of many lifestyle variables including how many hours of sleep obtained on a typical day. Researchers found that visually impaired students averaged 8.86 hours of sleep, with a standard deviation of 1.7 hours. Assume that the number of hours of sleep for these visually impaired students is normally distributed.

(a) What is the probability that a visually impaired student gets at most 6.2 hours of sleep? Express your answer as a percent rounded to 2 decimal places. e.g. 1.23% Do not include the % symbol in your answer.

Answer: %

(b) What is the probability that a visually impaired student gets between 6.9 and 7.84 hours of sleep? Express your answer as a percent rounded to 2 decimal places. e.g. 1.23% Do not include the % symbol in your answer.

Answer: %

(c) What is the probability that a visually impaired student gets at least 7.1 hours of sleep? Express your answer as a percent rounded to 2 decimal places. e.g. 1.23% Do not include the % symbol in your answer.

Answer: %

(d) What is the sleep time that cuts off the top 35.9% of sleep hours? Round your answer to 2 decimal places.

Answer: hours

(e) If 500 visually impaired students were studied, how many students would you expect to have sleep times of more than 7.84 hours? Round to the nearest whole number.

Answer: students

(f) A school district wants to give additional assistance to visually impaired students with sleep times at the first quartile and lower. What would be the maximum sleep time to be recommended for additional assistance? Round your answer to 2 decimal places.

Answer: hour

To solve these questions, we will use the Z-score formula:

Z = (X - μ) / σ
Where Z is the Z-score, X is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.

(a) To find the probability that a visually impaired student gets at most 6.2 hours of sleep, we need to calculate the Z-score for 6.2 and find the corresponding probability using a Z-table.
Z = (6.2 - 8.86) / 1.7 = -1.56
Using the Z-table, we can find that the probability corresponding to -1.56 is 0.0594. Multiplying by 100, the probability is 5.94%.

Answer: 5.94%

(b) To find the probability that a visually impaired student gets between 6.9 and 7.84 hours of sleep, we need to calculate the Z-scores for both values and find the difference between their probabilities.
Z1 = (6.9 - 8.86) / 1.7 = -1.15
Z2 = (7.84 - 8.86) / 1.7 = -0.6
Using the Z-table, we can find that the probability corresponding to -1.15 is 0.1251 and the probability corresponding to -0.6 is 0.2743. The difference between these two probabilities is 0.2743 - 0.1251 = 0.1492. Multiplying by 100, the probability is 14.92%.

Answer: 14.92%

(c) To find the probability that a visually impaired student gets at least 7.1 hours of sleep, we need to calculate the Z-score for 7.1 and find the corresponding probability using a Z-table.
Z = (7.1 - 8.86) / 1.7 = -1.04
Using the Z-table, we can find that the probability corresponding to -1.04 is 0.1492. Multiplying by 100, the probability is 14.92%.

Answer: 14.92%

(d) To find the sleep time that cuts off the top 35.9% of sleep hours, we need to find the Z-score corresponding to this probability using a Z-table.
The Z-score that corresponds to a probability of 35.9% is approximately 0.359.
Using the Z-score formula, we can solve for X:
0.359 = (X - 8.86) / 1.7
X - 8.86 = 0.359 * 1.7
X - 8.86 = 0.6113
X = 8.86 + 0.6113 = 9.4713
Rounding to 2 decimal places, the sleep time that cuts off the top 35.9% of sleep hours is 9.47 hours.

Answer: 9.47 hours

(e) To estimate the number of students out of 500 who would have sleep times of more than 7.84 hours, we need to find the Z-score for 7.84 and then find the corresponding percentile using a Z-table.
Z = (7.84 - 8.86) / 1.7 = -0.6
From the Z-table, we can find that the percentile corresponding to -0.6 is approximately 27.35%.
Out of 500 students, 27.35% would be 0.2735 * 500 = 136.75.
Rounding to the nearest whole number, we would expect approximately 137 students to have sleep times of more than 7.84 hours.

Answer: 137 students

(f) The first quartile represents the 25th percentile. To find the maximum sleep time to be recommended for additional assistance, we need to find the value that corresponds to the 25th percentile using a Z-table.
The Z-score that corresponds to the 25th percentile is approximately -0.674.
Using the Z-score formula, we can solve for X:
-0.674 = (X - 8.86) / 1.7
X - 8.86 = -0.674 * 1.7
X - 8.86 = -1.1448
X = 8.86 - 1.1448 = 7.7152
Rounding to 2 decimal places, the maximum sleep time to be recommended for additional assistance is 7.72 hours.

Answer: 7.72 hours

To answer these questions, we'll make use of the normal distribution and the properties of z-scores. A z-score measures how many standard deviations a particular value is from the mean in a normal distribution.

(a) To find the probability that a visually impaired student gets at most 6.2 hours of sleep, we need to convert 6.2 hours into a z-score and then find the corresponding probability.

The formula to calculate the z-score is: z = (x - μ) / σ

Where:
x = 6.2 hours
μ = mean of sleep hours = 8.86 hours
σ = standard deviation = 1.7 hours

Plugging in the values, we get:
z = (6.2 - 8.86) / 1.7 = -1.56

To find the probability, we can use a standard normal distribution table or calculator. We need to find the area to the left of the z-score -1.56. From the table/calculator, we find that the probability is 0.0594.

Converting this probability to a percentage and rounding to 2 decimal places, we get: 5.94% (answer to part a).

(b) To find the probability that a visually impaired student gets between 6.9 and 7.84 hours of sleep, we need to find the probabilities for both values separately and then calculate the difference.

Following the same process as in part (a), we find:
For 6.9 hours: z = (6.9 - 8.86) / 1.7 = -1.15
For 7.84 hours: z = (7.84 - 8.86) / 1.7 = -0.60

From the standard normal distribution table/calculator, we find the following probabilities:
For z = -1.15, probability = 0.1251
For z = -0.60, probability = 0.2743

The probability of getting between 6.9 and 7.84 hours is the difference between these probabilities: 0.2743 - 0.1251 = 0.1492.

Converting this probability to a percentage and rounding to 2 decimal places, we get: 14.92% (answer to part b).

(c) To find the probability that a visually impaired student gets at least 7.1 hours of sleep, we need to find the probability to the right of 7.1 hours.

Calculating the z-score for 7.1 hours:
z = (7.1 - 8.86) / 1.7 = -1.05

From the standard normal distribution table/calculator, we find the probability to the right of -1.05, which is 0.8549.

Converting this probability to a percentage and rounding to 2 decimal places, we get: 85.49% (answer to part c).

(d) To find the sleep time that cuts off the top 35.9% of sleep hours, we need to find the z-score corresponding to this percentile.

From the standard normal distribution table/calculator, we find the z-score that corresponds to a cumulative probability of 0.359, which is approximately 0.3476.

To find the corresponding sleep time, we use the z-score formula and solve for x:
0.3476 = (x - 8.86) / 1.7
Simplifying and solving for x, we get:
x = 0.3476 * 1.7 + 8.86

Rounding this result to 2 decimal places, we get: 8.53 hours (answer to part d).

(e) If 500 visually impaired students were studied, the number of students expected to have sleep times of more than 7.84 hours would depend on the proportion of students falling into that range. To find this, we need to find the probability of getting more than 7.84 hours and then multiply it by the total number of students (500).

Using the z-score formula, we find the z-score for 7.84 hours: z = (7.84 - 8.86) / 1.7 = -0.60

From the standard normal distribution table/calculator, we find the probability to the right of -0.60, which is 0.7257.

The number of students expected to have sleep times of more than 7.84 hours would be: 0.7257 * 500 = 362.85. Rounding to the nearest whole number, we get: 363 students (answer to part e).

(f) To find the maximum sleep time to be recommended for additional assistance for visually impaired students at the first quartile and lower, we need to find the value corresponding to the first quartile (25th percentile) in the distribution.

From the standard normal distribution table/calculator, we find the z-score that corresponds to a cumulative probability of 0.25, which is approximately -0.6745.

To find the corresponding sleep time, we use the z-score formula and solve for x:
-0.6745 = (x - 8.86) / 1.7
Simplifying and solving for x, we get:
x = -0.6745 * 1.7 + 8.86

Rounding this result to 2 decimal places, we get: 7.73 hours (answer to part f).

dadfadf

a-c. Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities of the Z scores. Multiply by 100 to get percentage.

d. Go to table first to find Z, then insert into equation above.

e. Same a, but multiply by 500.

f. first quartile = .2500. Use that Z score.