A health expert evaluates the sleeping patterns of adults. Each week she randomly selects 30 adults and calculates their average sleep time. Over many weeks, she finds that 5% of average sleep time is less than 9 hours and 5% of average sleep time is more than 9.4 hours. What are the mean and standard deviation (in hours) of sleep time for the population? (Round "Mean" to 1 decimal places and "standard deviation" to 3 decimal places.)
To find the mean and standard deviation of sleep time for the population, we can use the z-score formula.
Let's assume the mean sleep time is μ, and the standard deviation is σ. We need to find the values of μ and σ.
Given that 5% of average sleep time is less than 9 hours, we can convert this value to a z-score using the z-score formula:
z = (x - μ) / σ,
where x is the value of 9 hours.
So, we have:
(9 - μ) / σ = -1.645
Similarly, given that 5% of average sleep time is more than 9.4 hours, we can convert this value to a z-score using the z-score formula:
z = (x - μ) / σ,
where x is the value of 9.4 hours.
So, we have:
(9.4 - μ) / σ = 1.645
Now, we have a system of two equations with two variables:
(9 - μ) / σ = -1.645
(9.4 - μ) / σ = 1.645
To solve this system, we can isolate μ in both equations:
9 - μ = -1.645σ
9.4 - μ = 1.645σ
Now, we can solve for μ by subtracting the first equation from the second equation:
(9.4 - μ) - (9 - μ) = 1.645σ + 1.645σ
0.4 = 3.29σ
Simplifying, we have:
σ = 0.4 / 3.29
Now, we substitute the value of σ back into either of the original equations to solve for μ. Let's choose the first equation:
(9 - μ) / (0.4 / 3.29) = -1.645
Multiplying both sides of the equation by (0.4 / 3.29) and simplifying, we get:
9 - μ = -0.2188
Subtracting -0.2188 from both sides of the equation, we get:
μ = 9 + 0.2188 = 9.2188
So, the mean sleep time for the population is approximately 9.2 hours (rounded to 1 decimal place).
The standard deviation is given by:
σ = 0.4 / 3.29 = 0.1217
So, the standard deviation of sleep time for the population is approximately 0.122 hours (rounded to 3 decimal places).
To find the mean and standard deviation of sleep time for the population, we need to use the concept of z-scores and the standard normal distribution.
Let's assume that the average sleep time for the population follows a normal distribution.
Step 1: Find the z-score corresponding to the 5th percentile.
A z-score represents the number of standard deviations away from the mean. We want to find the z-score that corresponds to the 5th percentile, which is the value below which 5% of the data falls.
Using a standard normal distribution table or a calculator, we find that the z-score corresponding to the 5th percentile is approximately -1.645.
Step 2: Find the z-score corresponding to the 95th percentile.
Similarly, we want to find the z-score that corresponds to the 95th percentile, which is the value above which 5% of the data falls.
Using the same standard normal distribution table or calculator, we find that the z-score corresponding to the 95th percentile is approximately 1.645.
Step 3: Convert the z-scores to actual sleep time values.
To convert z-scores to actual sleep time values, we use the formula:
x = z * standard deviation + mean
For the 5th percentile:
9 hours = -1.645 * standard deviation + mean (equation 1)
For the 95th percentile:
9.4 hours = 1.645 * standard deviation + mean (equation 2)
Step 4: Solve the system of equations.
Solving equations 1 and 2 simultaneously will give us the values of the mean and standard deviation.
Subtracting equation 1 from equation 2:
9.4 hours - 9 hours = 1.645 * standard deviation + mean - (-1.645 * standard deviation + mean)
0.4 hours = 1.645 * standard deviation - (-1.645 * standard deviation)
0.4 hours = 3.29 * standard deviation
Dividing both sides by 3.29:
standard deviation = 0.4 / 3.29
standard deviation = 0.1217 (approximately)
Substituting the value of the standard deviation into equation 1 or 2, we can find the mean:
9 hours = -1.645 * 0.1217 + mean
mean = 9 + 1.645 * 0.1217
mean = 9.2 (approximately)
Therefore, the mean sleep time for the population is approximately 9.2 hours, and the standard deviation is approximately 0.1217 hours.