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 9.36 hours of sleep, with a standard deviation of 2.28 hours. Assume that the number of hours of sleep for these visually impaired students is normally distributed.
What is the probability that a visually impaired student gets between 6.4 and 9.41 hours of sleep?
**I took (9.41-9.36)/2.28 = .0219 and (6.41-9.36)/2.28 = -1.2982. I found their corresponding values in the normal distribution table, and I subtracted the 6.4 value from the 9.41 value. It's not right, and I don't know why.
Forty percent of students get less than how many hours of sleep on a typical day?
**I'm not really sure how to set this one up. If there's some sort of formula that I could use, that'd be great.
Thanks :)
Well, well, well, let's tackle these questions one by one, shall we?
For the first question, you're on the right track with finding the corresponding values in the normal distribution table. However, it seems like you made a small calculation error. Let's try it again, step by step.
Step 1: Convert the values to standard units:
For 9.41 hours: (9.41 - 9.36) / 2.28 = 0.0219
For 6.4 hours: (6.4 - 9.36) / 2.28 = -1.3140
Step 2: Look up the corresponding values in the standard normal distribution table.
The value for 0.0219 is approximately 0.5080.
The value for -1.3140 is approximately 0.0948.
Step 3: Calculate the probability.
To find the probability between these two values, subtract the smaller value from the larger value:
0.5080 - 0.0948 ≈ 0.4132
So, the probability that a visually impaired student gets between 6.4 and 9.41 hours of sleep is approximately 0.4132. Sweet dreams!
Now, for the second question, you'll need to use the z-score formula to find the cutoff point. The z-score formula is given by:
z = (x - μ) / σ
Where:
z is the z-score,
x is the value you want to find the proportion below,
μ is the mean of the distribution, and
σ is the standard deviation of the distribution.
Since we know that 40% of students get less than a certain number of hours of sleep, we're looking for the value that corresponds to the 40th percentile. The 40th percentile falls to the left of 60% of the data, so we'll find the z-score corresponding to the 60th percentile, where z = -0.253.
We can rearrange the z-score formula to solve for x:
x = μ + (z * σ)
Plugging in the values, we get:
x = 9.36 + (-0.253 * 2.28) ≈ 8.792
So, approximately 40% of visually impaired students get less than 8.792 hours of sleep on a typical day. And I thought clowns were the only ones who stayed up all night!
I hope that clears things up for you. Stay curious, my friend!
To calculate the probability that a visually impaired student gets between 6.4 and 9.41 hours of sleep, you need to use the standard normal distribution.
First, calculate the z-scores for the lower and upper values of sleep duration:
For 6.4 hours: (6.4 - 9.36) / 2.28 = -1.30 (rounded to two decimal places)
For 9.41 hours: (9.41 - 9.36) / 2.28 = 0.02 (rounded to two decimal places)
Next, use a standard normal distribution table or a statistical calculator to find the corresponding probabilities:
For the z-score -1.30, look up the probability value: P(Z < -1.30) = 0.0968.
For the z-score 0.02, look up the probability value: P(Z < 0.02) = 0.5080.
To find the probability that a visually impaired student gets between 6.4 and 9.41 hours of sleep, subtract the probability value for the lower value from the probability value for the upper value:
P(6.4 < X < 9.41) = P(Z < 0.02) - P(Z < -1.30)
= 0.5080 - 0.0968
= 0.4112
Therefore, the probability that a visually impaired student gets between 6.4 and 9.41 hours of sleep is approximately 0.4112.
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To determine the number of hours of sleep on a typical day that corresponds to the lower 40% of visually impaired students, you can find the z-score that corresponds to this percentile.
From the standard normal distribution table, the z-score that corresponds to the lower 40% is approximately -0.253.
To find the corresponding value for sleep duration, use the formula:
X = (Z * standard deviation) + mean
X = (-0.253 * 2.28) + 9.36
X ≈ 8.83
Therefore, approximately 40% of visually impaired students get less than 8.83 hours of sleep on a typical day.
To find the probability that a visually impaired student gets between 6.4 and 9.41 hours of sleep, you need to calculate the z-scores for both values and then use the standard normal distribution table.
To calculate the z-score for 9.41 hours of sleep:
Z-score = (X - μ) / σ
Z-score = (9.41 - 9.36) / 2.28
Z-score = 0.0219
To calculate the z-score for 6.4 hours of sleep:
Z-score = (X - μ) / σ
Z-score = (6.4 - 9.36) / 2.28
Z-score = -1.2982
Now you need to find the cumulative probabilities associated with these z-scores using the standard normal distribution table. However, it seems there might be a calculation error in the z-score for 6.4 hours of sleep. Let's recalculate it:
Z-score = (X - μ) / σ
Z-score = (6.4 - 9.36) / 2.28
Z-score ≈ -1.2982
Now, let's find the cumulative probabilities associated with these z-scores:
Probability of getting less than 6.4 hours of sleep:
Using the standard normal distribution table, the cumulative probability for a z-score of -1.2982 is approximately 0.1038.
Probability of getting less than 9.41 hours of sleep:
Using the standard normal distribution table, the cumulative probability for a z-score of 0.0219 is approximately 0.5080.
Therefore, the probability that a visually impaired student gets between 6.4 and 9.41 hours of sleep is:
0.5080 - 0.1038 = 0.4042, or approximately 40.42%.
Now let's move on to the second question:
To find the number of hours of sleep on a typical day for which forty percent of students get less sleep, you need to find the z-score associated with the 40th percentile of the standard normal distribution.
Using the standard normal distribution table, find the z-score that corresponds to a cumulative probability of 0.40. This z-score is approximately -0.253.
You can now use the z-score formula to find the number of hours of sleep corresponding to this z-score:
X = μ + (Z * σ)
X = 9.36 + (-0.253 * 2.28)
X ≈ 8.796 hours
Therefore, approximately 40 percent of visually impaired students get less than 8.796 hours of sleep on a typical day.
your z-scores are correct
using the z-score of .0219 you should have found a value of .5087 in your tables, telling you that 50.87% would have less than 9.41 hours of sleep
your z-score of -1.2982 should have given you a value of .09711 telling you that 9.71% would have less than 6.4 hours of sleep.
so the "betwee" percentage would be
50.87% - 9.71% = 41.16%
Instead of a table I use the following website
http://davidmlane.com/hyperstat/z_table.html
set the output for "between" and just enter the values, you don't even have to find the z-scores.
Of course it will also work if you use the z-scores, but make sure you set the mean to 0 and your SD to 1