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.)
To solve this problem, we will first need to convert the normally distributed random variable X to a standard normal distribution using the z-score transformation. Let's denote the standard normal random variable as Z.
The z-score transformation can be calculated using the formula:
Z = (X - μ) / σ
Where:
- X is the observed value of the random variable
- μ is the mean of the random variable X
- σ is the standard deviation of the random variable X
In this case, X has a mean of 15 and a standard deviation of 0.25. So the z-score transformation becomes:
Z = (X - 15) / 0.25
Now, we need to find the corresponding values of Z, denoted as zL and zR, that satisfy P(zL < Z < zR) = 0.80.
Since the standard normal distribution is symmetric, the area under the curve between zL and zR is equal to half the desired probability, which is 0.80 / 2 = 0.40.
Using the standard normal distribution table or software, we can find the z-scores corresponding to this probability of 0.40.
The z-score corresponding to a cumulative probability of 0.40 on the left side of the curve is -0.253. To find zL, we calculate:
zL = -0.253
To find zR, we use the symmetric property of the standard normal distribution. Thus, zR is the negation of zL:
zR = -(-0.253) = 0.253
Now, we can convert these z-scores back to the original X values using the z-score transformation equation:
X = Z * σ + μ
Plugging in the values, we get:
xL = zL * σ + μ = -0.253 * 0.25 + 15 = 14.93675
xR = zR * σ + μ = 0.253 * 0.25 + 15 = 15.06325
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 14.937 and 15.063, respectively.
To find the values xL and xR of X that are symmetrically located with respect to the mean and satisfy P(xL < X < xR) = 0.80, we need to first solve the corresponding problem for Z.
Z is a standard normal random variable with mean 0 and standard deviation 1.
Since X is normally distributed with mean 15 and standard deviation 0.25, we can standardize X to Z using the formula:
Z = (X - μ) / σ
where μ is the mean and σ is the standard deviation.
Substituting the given values, we have:
Z = (X - 15) / 0.25
Now, we can rewrite the inequality P(xL < X < xR) = 0.80 in terms of Z:
P(xL < X < xR) = P((xL - 15) / 0.25 < (X - 15) / 0.25 < (xR - 15) / 0.25) = 0.80
Since we want the values xL and xR to be symmetrically located with respect to the mean of X, we can rewrite the above inequality as:
P(-a < Z < a) = 0.80
where a = (15 - xL) / 0.25
To find the value of a, we need to determine the corresponding z-scores for the 0.80 probability. Using the standard normal distribution table or a calculator, we can find that the z-score that corresponds to a 0.80 probability is approximately 0.842.
So,
0.842 = (15 - xL) / 0.25
Solving for xL:
0.842 * 0.25 = 15 - xL
0.2105 = 15 - xL
xL = 15 - 0.2105
xL ≈ 14.7895
Similarly, we can find xR:
xR = 15 + 0.2105
xR ≈ 15.2105
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 14.7895 and 15.2105, respectively.