Answer: a) 0.5328 b) 0.9772
Step-by-step explanation:
Given : Resistors to be used in a circuit have average resistance 200 ohms and standard deviation 10 ohms.
[tex]\mu=200[/tex] and [tex]\sigma=10[/tex]
We assume that the resistance in circuits are normally distributed.
a) Let x denotes the average resistance of the circuit.
Sample size : n= 25
Then, the probability that the average resistance for the 25 resistors is between 199 and 202 ohms :-
[tex]P(199<x<200)=P(\dfrac{199-200}{\dfrac{10}{\sqrt{25}}}<\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{202-200}{\dfrac{10}{\sqrt{25}}})\\\\=P(-0.5<z<1)\ \ [\because z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=P(z<1)-P(z<-0.5)\\\\=P(z<1)-(1-P(z<0.5))\ \ [\because\ P(Z<-z)=1-P(Z<z)]\\\\=0.8413-(1-0.6915)\ \ [\text{By z-table}]\\\\=0.5328[/tex]
b) Total resistors = 25
Let Z be the total resistance of 25 resistors.
To find P(Z≤5100 ohms) , first we find the mean and variance for Z.
Mean= E(Y) = E(25 X)=25 E(X)=25(200)= 5000 ohm
[tex]Var(Y)=Var(25\ X)=25^2(\dfrac{\sigma^2}{n})=25^2\dfrac{(10)^2}{25}=2500[/tex]
The probability that the total resistance does not exceed 5100 ohms will be :
[tex]P(Y\leq5000)=P(\dfrac{Y-\mu}{\sqrt{Var(Y)}}<\dfrac{5100-5000}{\sqrt{2500}})\\\\=P(z\leq2)=0.9772\ \ [\text{By z-table}][/tex]
Hence, the probability that the total resistance does not exceed 5100 ohms = 0.9772
