A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 121.2-cm and a standard deviation of 1.8-cm.
Suppose a rod is chosen at random from all the rods produced by the company. There is a 23% probability that the rod is longer than:
Enter your answer as a number accurate to 1 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion in the smaller area (.23) and its Z score. Use the equation below.
Z = (score-mean)/SD
To find the length of the rod that has a 23% probability of being longer, we need to determine the z-score that corresponds to this probability in a standard normal distribution.
The z-score can be calculated using the following formula:
z = (x - μ) / σ
where:
z is the z-score
x is the value we want to find the probability for (in this case, the length of the rod)
μ is the mean of the distribution (121.2 cm)
σ is the standard deviation of the distribution (1.8 cm)
Step 1: Convert the probability to a z-score.
To find the z-score corresponding to a 23% probability, we need to find the z-score that has an area of 1 - 0.23 = 0.77 to its left. This corresponds to the z-score that marks the 77th percentile.
Step 2: Look up the z-score in a standard normal distribution table or use a calculator.
Using a standard normal distribution table or a calculator, we find that the z-score for the 77th percentile is approximately 0.60 (rounded to 2 decimal places).
Step 3: Solve for x (the length of the rod).
Now that we have the z-score, we can rearrange the formula to solve for x:
x = z * σ + μ
x = 0.60 * 1.8 + 121.2
x ≈ 122.08
Therefore, the length of the rod that has a 23% probability of being longer is approximately 122.1 cm (rounded to 1 decimal place).