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Answer :
About 19.74% of resistors are expected to have a resistance between 99.5 kΩ and 100.5 kΩ.
Firstly, we know that the mean resistance value is 100 kΩ and the standard deviation is 2 kΩ. We want to find the percentage of resistors that have a resistance between 99.5 kΩ and 100.5 kΩ.
To perform this calculation, we need to convert our values into Z-scores, which show us how many standard deviations from the mean our data points are. We calculate the Z-score for the lower and upper bounds of the resistance range.
For the lower bound (99.5 kΩ), the Z-score is calculated by subtracting the mean resistance value from the lower bound, and dividing by the standard deviation, which gives -0.25. This means that the lower bound is 0.25 standard deviations below the mean.
For the upper bound (100.5 kΩ), the Z-score is calculated the same way, which gives 0.25. This means that the upper bound is 0.25 standard deviations above the mean.
Now we need to calculate the probability that a randomly chosen resistor falls within this range. We do so by looking up these Z-scores on the Z-table or by using the cumulative distribution function for the normal distribution.
In our case, we subtract the value for the lower bound from the value for the upper bound (scipy.stats.norm.cdf(0.25) - scipy.stats.norm.cdf(-0.25)), which comes to approximately 0.19741.
Finally to get the percentage, we multiply this probability by 100, which gives about 19.74%.
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