To find the value of k, we need to determine the interval around the mean of the normal distribution X that contains 70% of the probability.
Given that X follows a normal distribution with a mean (μ) of 4 and a standard deviation (σ) of 2.3, we want to find the value of k such that Pr[(5-k) < X < (5+k)] = 0.7.
We can convert the interval (5-k) to (5+k) into standard units using the formula z = (x - μ) / σ, where z is the standard score.
So, the z-scores for the lower and upper bounds of the interval are:
z_lower = [(5 - k) - 4] / 2.3
z_upper = [(5 + k) - 4] / 2.3
Using the properties of the standard normal distribution, we can find the area between these two z-scores.
Pr[(5-k) < X < (5+k)] = Pr[z_lower < Z < z_upper] = 0.7
To find the corresponding z-scores for the cumulative probability of 0.7, we can use a standard normal distribution table or a calculator. The z-scores that enclose 70% of the area under the standard normal curve are approximately -0.524 and 0.524.
Now, equating the z-scores to the corresponding expressions:
-0.524 = [(5 - k) - 4] / 2.3
0.524 = [(5 + k) - 4] / 2.3
Solving these equations will give us the value of k.
-0.524 * 2.3 = 5 - k - 4
1.2068 = 1 - k
k = 1 - 1.2068
k ≈ -0.2068
0.524 * 2.3 = 5 + k - 4
1.2032 = k + 1
k ≈ 1.2032 - 1
k ≈ 0.2032
Therefore, the value of k is approximately 0.2032.