A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 139.1-cm and a standard deviation of 1.1-cm.
Find the probability that the length of a randomly selected steel rod is between 138.3-cm and 142.2-cm.
To find the probability that the length of a randomly selected steel rod is between 138.3-cm and 142.2-cm, we need to find the area under the normal distribution curve between these two values.
Step 1: Calculate the z-scores for the given lengths.
A z-score, also known as a standard score, measures how many standard deviations a value is from the mean. We can calculate the z-scores using the formula:
z = (x - μ) / σ
Where:
- x is the given length,
- μ is the mean of the distribution (139.1 cm), and
- σ is the standard deviation of the distribution (1.1 cm).
For 138.3 cm:
z1 = (138.3 - 139.1) / 1.1
For 142.2 cm:
z2 = (142.2 - 139.1) / 1.1
Step 2: Look up the z-scores in the z-table.
A z-table is a table that provides the area under the normal distribution curve up to a given z-score. By looking up the z-scores from Step 1 in the z-table, we can determine the corresponding areas.
Let's assume that the z-scores rounded to two decimal places are z1 = -0.73 and z2 = 2.82.
Step 3: Calculate the probability.
To find the probability that the length of a randomly selected steel rod is between 138.3-cm and 142.2-cm, we need to find the area between the z-scores from Step 2.
P(138.3 cm < X < 142.2 cm) = P(z1 < Z < z2)
Using the z-table, we can determine the area to the left of z1 (0.2327) and the area to the left of z2 (0.9978). To find the area between these two z-scores, we subtract the smaller area from the larger area:
P(z1 < Z < z2) = P(Z < z2) - P(Z < z1)
P(z1 < Z < z2) = 0.9978 - 0.2327 (round off the values to decimal places)
P(z1 < Z < z2) = 0.7651
So, the probability that the length of a randomly selected steel rod is between 138.3-cm and 142.2-cm is approximately 0.7651 (or 76.51%).