According to the Sleep Foundation, the average night's sleep is 6.8 hours (Fortune, March 20, 2006). Assume the standard deviation is .7 hours and that the probability distribution is normal.
a. What is the probability that a randomly selected person sleeps more than 8 hours (to 4 decimals)?
b. What is the probability that a randomly selected person sleeps 6 hours or less (to 4 decimals)?
c. Doctors suggest getting between 7 and 9 hours of sleep each night. What percentage of the population gets this much sleep (to the nearest whole number)?
37) According to the Sleep Foundation, the average night's sleep is 6.8 hours (Fortune, March 20, 2006). Assume the standard deviation is .7 hours and that the probability distribution is normal.
a. What is the probability that a randomly selected person sleeps more than 8 hours (to 4 decimals)?
P( x > 8)
z = ( x - μ ) / σ
z = ( 8 - 6.8 ) / .7
z = 1.714
P(z > 1.714) = 1-0.9564 = 0.0436
b. What is the probability that a randomly selected person sleeps 6 hours or less (to 4 decimals)?
P ( x < 6)
z = ( x - μ ) / σ
z = ( 6 - 6.8 ) / .7
z = -1.14
P( z <-1.14) = 0.1271
c. Doctors suggest getting between 7 and 9 hours of sleep each night. What percentage of the population gets this much sleep (to the nearest whole number)?
P( 7 < x < 9)
z = ( x - μ ) / σ
z = ( 7 - 6.8 ) / .7
z = 0.29
z = ( x - μ ) / σ
z = ( 9 - 6.8 ) / .7
z = 3.14
P ( 0.29 < z < 3.14) = 0.9992 - 0.6141 = 0.3851
38.51%
A-
Z= 8-6.8/6=1.2/6=0.2
B-
Z= 6-6.8/6=0.8/6= 0.133
C-
Z= 7-6.8/6= 0.2/6= 0.333
Z= 9-6.8/6=2.2/6=0.366
Sorry, but there are a couple of errors in your calculations.
A-
To find the z-score, we need to use the formula:
z = (x - μ) / σ
So,
z = (8 - 6.8) / 0.7 = 1.71
Using the standard normal distribution table or calculator, we find that the probability of z being greater than 1.71 is 0.0438. Therefore, the probability of a randomly selected person sleeping more than 8 hours is 0.0438, or 4.38% (rounded to two decimal places).
B-
To find the probability of a randomly selected person sleeping 6 hours or less, we need to find the area under the normal distribution curve to the left of 6.
z = (6 - 6.8) / 0.7 = -1.14
Using the standard normal distribution table or calculator, we find that the probability of z being less than -1.14 is 0.1271. Therefore, the probability of a randomly selected person sleeping 6 hours or less is 0.1271, or 12.71% (rounded to two decimal places).
C-
To find the probability of a randomly selected person sleeping between 7 and 9 hours, we need to find the area under the normal distribution curve between the z-scores corresponding to 7 and 9.
z1 = (7 - 6.8) / 0.7 = 0.286
z2 = (9 - 6.8) / 0.7 = 3.14
Using the standard normal distribution table or calculator, we find that the probability of z being between 0.286 and 3.14 is 0.4966. Therefore, the probability of a randomly selected person sleeping between 7 and 9 hours is 0.4966, or 49.66% (rounded to two decimal places).
a. Well, some people reread a book before bed, but apparently, a randomly selected person has a 0.0228 probability of sleeping more than 8 hours. That's definitely more time for dreaming about winning the lottery!
b. Ah, the early birds! The probability of a randomly selected person sleeping 6 hours or less is 0.1056. It seems like these people are always awake before the sun, ready to seize the day or maybe just indulge in some breakfast cereal.
c. Ah, the magical range of 7 to 9 hours of sleep per night recommended by doctors. Approximately 68% of the population manages to find their way into that sweet spot. So that's a whole lot of people who might still have a chance at catching a unicorn in their dreams.
To answer these questions, we can use the Z-score formula, which allows us to standardize the values by converting them into Z-scores. A Z-score tells us how many standard deviations a given value is away from the mean.
First, let's calculate the Z-score for each question. The formula for calculating the Z-score is:
Z = (X - μ) / σ
where:
Z = Z-score
X = given value
μ = mean
σ = standard deviation
a. Probability of sleeping more than 8 hours:
Given X = 8 hours, μ = 6.8 hours, and σ = 0.7 hours
Using the Z-score formula, we get:
Z = (8 - 6.8) / 0.7
To find the probability, we need to find the area to the right of the Z-score (8 hours is greater than the mean, so it is on the right side of the distribution).
b. Probability of sleeping 6 hours or less:
Given X = 6 hours, μ = 6.8 hours, and σ = 0.7 hours
Using the Z-score formula, we get:
Z = (6 - 6.8) / 0.7
To find the probability, we need to find the area to the left of the Z-score (6 hours is less than the mean, so it is on the left side of the distribution).
c. Percentage of the population getting between 7 and 9 hours of sleep:
To find this percentage, we need to find the area between the Z-scores corresponding to 7 and 9 hours.
We will use a standard normal distribution table or a Z-table to look up the probabilities corresponding to the calculated Z-scores. The Z-table provides the cumulative probability up to a given Z-score. It is usually given for Z-scores rounding to two decimal places.
Let's calculate these probabilities step by step.
a. Probability of sleeping more than 8 hours:
Z = (8 - 6.8) / 0.7 = 1.71 (rounded to two decimal places)
Using the Z-table, we can find the probability corresponding to a Z-score of 1.71. The table gives us the area to the left of the Z-score. Since we want the area to the right, we subtract the left side probability from 1:
P(X > 8) = 1 - P(Z < 1.71)
b. Probability of sleeping 6 hours or less:
Z = (6 - 6.8) / 0.7 = -1.14 (rounded to two decimal places)
Using the Z-table, we can find the probability corresponding to a Z-score of -1.14. The table gives us the area to the left of the Z-score:
P(X ≤ 6) = P(Z < -1.14)
c. Percentage of the population getting between 7 and 9 hours of sleep:
To find the percentage, we need to calculate the area between the Z-scores corresponding to 7 and 9 hours. Let's calculate the Z-scores for 7 and 9 hours.
For 7 hours:
Z1 = (7 - 6.8) / 0.7
For 9 hours:
Z2 = (9 - 6.8) / 0.7
Using the Z-table, we can find the probabilities corresponding to these Z-scores. To find the area between the two Z-scores, we subtract the left side probability of Z1 from the left side probability of Z2:
P(7 ≤ X ≤ 9) = P(Z1 ≤ Z ≤ Z2)
Once we have the probabilities, we can convert them to percentages by multiplying by 100.
Remember, these calculations assume a normal distribution for sleep duration and the given mean and standard deviation.