Answer:
= 99 Ω
= 2.3094 Ω
P(98<R<102) = 0.5696
Step-by-step explanation:
The mean resistance is the average of edge values of interval.
Hence,
The mean resistance,
= 99 Ω
To find the standard deviation of resistance, we need to find variance first.

Hence,
The standard deviation of resistance,
= 2.3094 Ω
To calculate the probability that resistance is between 98 Ω and 102 Ω, we need to find Normal Distributions.


From the Z-table, P(98<R<102) = 0.9032 - 0.3336 = 0.5696