Let x be a random variable that represents the length of time it takes a student to complete a take-home exam in Dr. Larson’s psychology class. After interviewing many students, it was found that x has an approximately normal distribution with mean µ = 5.2 hours and standard deviation σ = 1.8 hours. Convert the x interval x ≥ 9.7 to a standard z interval.
Z = (score-mean)/SD
Z = (9.7-5.2)/1.8 = ?
2.5
To convert the x interval x ≥ 9.7 to a standard z interval, we need to use the formula for standardizing a random variable:
z = (x - µ) / σ
where z is the standard score, x is the raw score, µ is the population mean, and σ is the population standard deviation.
In this case, we have x ≥ 9.7. To convert this to a z interval, we will use the upper bound of 9.7.
Using the formula, we can calculate the z-score for x = 9.7:
z = (9.7 - 5.2) / 1.8
z = 4.5 / 1.8
z = 2.5
So, the z-score for x = 9.7 is 2.5.
Now we have a z-score of 2.5. To find the corresponding x value, we need to use the inverse of the standardizing formula:
x = µ + (z * σ)
Plugging in the values, we get:
x = 5.2 + (2.5 * 1.8)
x = 5.2 + 4.5
x = 9.7
So, the x interval x ≥ 9.7 converts to the z interval z ≥ 2.5.
To convert the x interval x ≥ 9.7 to a standard z interval, we need to standardize the values using the given mean and standard deviation. The formula for standardizing a value x to a z-score is:
z = (x - µ) / σ
Given:
µ = 5.2 hours
σ = 1.8 hours
x = 9.7 hours
To find the z-score for x = 9.7, we substitute these values into the formula:
z = (9.7 - 5.2) / 1.8
z = 4.5 / 1.8
z ≈ 2.5
So, the z-score for x = 9.7 is approximately 2.5.
To convert the interval x ≥ 9.7 to a standard z interval, we can use the z-score.
The z interval for this x interval can be written as:
z ≥ 2.5
Therefore, the converted interval from x ≥ 9.7 to a standard z interval is z ≥ 2.5.