A machine produces parts with a mean diameter of 5 centimeters and a standard deviation of 0.05 centimeters. What is the probability that a randomly selected part has a diameter between 4.938 centimeters and 4.984 centimeters?
To find the probability that a randomly selected part has a diameter between 4.938 cm and 4.984 cm, we first need to standardize the values using the z-score formula:
z = (X - μ) / σ
For the lower bound:
z1 = (4.938 - 5) / 0.05 = -1.24
For the upper bound:
z2 = (4.984 - 5) / 0.05 = -0.32
Next, we can find the corresponding probabilities for these z-scores using a standard normal distribution table or a calculator.
The probability of a z-score of -1.24 or lower is 0.1093, and the probability of a z-score of -0.32 or lower is 0.3745.
To find the probability of a randomly selected part having a diameter between 4.938 cm and 4.984 cm, we subtract the probability of the z-score being less than 4.938 cm from the z-score being less than 4.984 cm:
P(4.938 < X < 4.984) = P(-1.24 < z < -0.32) = P(z < -0.32) - P(z < -1.24) = 0.3745 - 0.1093 = 0.2652
Therefore, the probability that a randomly selected part has a diameter between 4.938 cm and 4.984 cm is 0.2652, or 26.52%.