X is a normally distributed random variable X with mean 15 and standard deviation 0.25. Find the values xL and xR of X that are symmetrically located with respect to the mean of X and satisfy P(xL < X < xR) = 0.80. (Hint. First solve the corresponding problem for Z.)
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (±.40) and the corresponding Z scores. Insert in above equation to get the raw scores.
15.32
To solve this problem, we can first convert X into a standard normal distribution by using the formula Z = (X - mean) / standard deviation.
Given that X has a mean of 15 and a standard deviation of 0.25, we can calculate Z using the formula Z = (X - 15) / 0.25.
Now, let's solve the problem for Z. We need to find the values for zL and zR that are symmetrically located about the mean of Z and satisfy P(zL < Z < zR) = 0.80.
To find these values, we need to consult the standard normal distribution table. From the table, we know that the area to the left of zL is 0.10, and the area to the left of zR is 0.90. Since the total area under the standard normal distribution curve is 1, we can subtract these values from 1 to find the areas to the right of zL and zR, respectively.
The area to the right of zL is 1 - 0.10 = 0.90, and the area to the right of zR is 1 - 0.90 = 0.10. Since the standard normal distribution is symmetric, these two areas are equal.
Now that we have these areas, we can use the standard normal distribution table (or a calculator) to find the z-scores corresponding to these areas. From the table, we find that the z-score corresponding to an area of 0.90 is approximately 1.28, and the z-score corresponding to an area of 0.10 is approximately -1.28.
Finally, we convert these z-scores back to X values using the formula X = Z * standard deviation + mean.
xL = -1.28 * 0.25 + 15 = 14.68
xR = 1.28 * 0.25 + 15 = 15.32
Therefore, the values xL and xR of X that are symmetrically located with respect to the mean of X and satisfy P(xL < X < xR) = 0.80 are 14.68 and 15.32, respectively.
To solve this problem, we first need to standardize the random variable X by converting it to a standard normal random variable Z. We can do this using the formula:
Z = (X - μ) / σ
where Z is the standard normal random variable, X is the normally distributed random variable, μ is the mean of X, and σ is the standard deviation of X.
In this case, X has a mean (μ) of 15 and a standard deviation (σ) of 0.25. So, we have:
Z = (X - 15) / 0.25
Now, we want to find the values xL and xR of X such that P(xL < X < xR) = 0.80.
To find the corresponding values of Z, we use the standard normal distribution table or a calculator to find the value of Z that corresponds to a cumulative probability of 0.80. Let's call this value Zc.
Now, using the formula for standardization, we have:
Zc = (xL - μ) / σ
Since the values xL and xR are symmetrically located with respect to the mean of X, we can write:
-xL = xR
Substituting these into the standardization formula, we have:
Zc = (-xL - 15) / 0.25
Simplifying this equation, we get:
-0.25Zc - 15 = xL
And since xL = -xR:
xR = 0.25Zc + 15
Now we have two equations:
-0.25Zc - 15 = xL
0.25Zc + 15 = xR
To solve these equations, we need to find the value of Zc that corresponds to a cumulative probability of 0.80 in the standard normal distribution table or a calculator.
Once we have Zc, we can substitute it into the equations above to find the values of xL and xR.
For example, if we find that Zc = 1.2816 (corresponding to a cumulative probability of 0.80), we can substitute it into the equations:
xL = -0.25(1.2816) - 15 = -15.3204
xR = 0.25(1.2816) + 15 = 15.3204
Therefore, the values xL and xR of X that are symmetrically located with respect to the mean of X and satisfy P(xL < X < xR) = 0.80 are approximately -15.3204 and 15.3204, respectively.