Answer:
To find the probability that a single randomly selected value is between 28.4 and 31.9 from a normal distribution with μ = 28.1 and σ = 54.1, we can use the standard normal distribution formula and then adjust for the given mean and standard deviation.
The formula for the standard normal distribution is:
\[ P(a < X < b) = Φ(b) - Φ(a) \]
where Φ(z) represents the cumulative distribution function (CDF) of the standard normal distribution.
First, let's standardize the values 28.4 and 31.9 using the formula:
\[ z = \frac{x - μ}{σ} \]
For 28.4:
\[ z_1 = \frac{28.4 - 28.1}{54.1} \]
For 31.9:
\[ z_2 = \frac{31.9 - 28.1}{54.1} \]
Now we'll look up the standard normal probabilities associated with these z-scores using a standard normal distribution table or calculator:
\[ P(Z < z_2) - P(Z < z_1) \]
Calculate \( z_1 \) and \( z_2 \) accurately:
\[ z_1 = \frac{0.3}{54.1} \]
\[ z_2 = \frac{3.8}{54.1} \]
Now calculate the probabilities:
\[ P(28.4 < X < 31.9) = P(Z < z_2) - P(Z < z_1) \]
Look up the standard normal probabilities for \( z_1 \) and \( z_2 \) in a table or calculator. Let's assume you find:
\[ P(Z < z_1) ≈ 0.2942 \]
\[ P(Z < z_2) ≈ 0.6423 \]
Now calculate the final probability:
\[ P(28.4 < X < 31.9) = 0.6423 - 0.2942 ≈ 0.3481 \]
So, \( P(28.4 < X < 31.9) ≈ 0.3481 \), which means there's approximately a 34.81% probability that a single randomly selected value from this distribution falls between 28.4 and 31.9.