The mean length of 200 diodes is 20.04mm, with a standard deviation of 0.02mm. What is the probability that a selected diode would have a length less than 20.01mm
20.01 mm is 0.03 mm below the mean ... this is 1.5 standard deviations
use a z-score table to find the portion of the population below -1.5 s.d.
should be around .07
20.01 is 1.5σ below the mean
So look that up in your Z table.
To find the probability that a selected diode would have a length less than 20.01mm, we need to use the concept of standard deviation and Z-scores.
First, let's calculate the Z-score for the value 20.01mm using the formula:
Z = (X - μ) / σ
Where:
Z = Z-score
X = Value we are interested in (20.01mm)
μ = Mean length of the diodes (20.04mm)
σ = Standard deviation (0.02mm)
Substituting the values into the formula:
Z = (20.01 - 20.04) / 0.02
Z = -0.03 / 0.02
Z = -1.5
The Z-score represents the number of standard deviations an observation is away from the mean. In this case, the value 20.01mm is 1.5 standard deviations below the mean.
Next, we need to find the probability associated with this Z-score. We can use a Z-table or a statistical calculator to find this value.
Looking up -1.5 in the Z-table, we find that the corresponding probability is approximately 0.0668 (or 6.68%).
Therefore, the probability that a selected diode would have a length less than 20.01mm is approximately 0.0668 or 6.68%.