Question

Suppose the force acting on a column that helps to support a building is a normally distributed random variable X with mean value 13.0 kips and standard deviation 1.50 kips. Compute the following probabilities by standardizing and then using a standard normal curve table from the Appendix Tables or SALT. (Round your answers to four decimal places.) USE SALT

(a) P(X ≤ 13)
(b) P(X ≤ 14.5)
(c) P(X≥ 5.5)
(d) P(11 ≤ x ≤ 16)
(e) P(|X 13| ≤ 2)

Answer 1

To compute the probabilities for the normally distributed force on a column, we can standardize the given values and use a Z-table. Values are converted using the formula Z = (X - mean) / standard deviation, and then the standard normal probabilities are looked up in the table. Probabilities for specific ranges are found by subtracting the probabilities of the relevant Z-values.

Step-by-step explanation:

To calculate the given probabilities for the normally distributed random variable X, with a mean value of 13.0 kips and standard deviation of 1.50 kips, we'll use standardization to convert X to the standard normal variable Z, and then use a standard normal distribution table (Z-table).

Step-by-Step Solutions:

1. To find P(X ≤ 13), we calculate Z for X=13:
   Z = (X - mean) / (standard deviation) = (13 - 13) / 1.5 = 0. Then P(X ≤ 13) is the probability that Z is less than or equal to 0, which is 0.5000.
2. For P(X ≤ 14.5), we find Z for X=14.5: Z = (14.5 - 13) / 1.5 ≈ 1. Then look up Z=1 in the Z-table to find the probability, which is about 0.8413.
3. To determine P(X ≥ 5.5), find Z for X=5.5: Z = (5.5 - 13) / 1.5 ≈ -5. Then use the Z-table to find P(Z ≤ -5), which is nearly 0, so P(X ≥ 5.5) ≈ 1 (since it's the complement).
4. P(11 ≤ X ≤ 16) is found by standardizing both values: Z for X=11 is (11 - 13) / 1.5 ≈ -1.33, and for X=16 it is (16 - 13) / 1.5 ≈ 2.00. Use the Z-table to find the corresponding probabilities and subtract them to get the result, which is approximately P(Z ≤ 2) - P(Z ≤ -1.33), equating to about 0.9772 - 0.0918 = 0.8854.
5. For P(|X - 13| ≤ 2), standardize the values X = 11 and X = 15 and find the probability between those Z-scores: P(-1.33 ≤ Z ≤ 1.33) ≈ P(Z ≤ 1.33) - P(Z ≤ -1.33), which gives about 0.9082 - 0.0918 = 0.8164.

Note: For parts (c) and (d), the exact probabilities depend on the precision of the standard normal distribution table used and the method of interpolation.