If a certain machine makes electrical resistors having a mean resistance of 40 ohms and a standard deviation of 2 ohms, what is the probability that a random sample of 36 of these resistors will have a combined resistance of more than 1458 ohms?

Respuesta :

Answer:0.0668

Step-by-step explanation:

Given

mean of sample [tex]\mu =40 [/tex]

Standard deviation [tex]\sigma =2[/tex]

For [tex]n=36[/tex] sample

Random variable [tex]Z=\frac{X-\mu }{\frac{\sigma }{\sqrt{n}}}[/tex]

For combined resistance of 1458 ohm mean resistance of [tex]\mu =\frac{1458}{36}=40.5[/tex]

[tex]P(40.5<X)=P(\frac{40.5-\mu }{\frac{2}{\sqrt{36}}}<\frac{X-\mu }{\frac{\sigma }{\sqrt{n}}})[/tex]

[tex]=P(\frac{40.5-40}{\frac{1}{3}}<Z)[/tex]

[tex]=P(1.5<Z)[/tex]

[tex]=1-P(Z\leq 1.5)[/tex]

[tex]=1-0.9332[/tex]

[tex]=0.0668[/tex]            

The probability is 0.0668

We have given that the mean resistance =40 ohms

stranded deviation =2 ohms

n=36

what is the formula for random variable?

[tex]Z=\frac{x-\mu }{\frac{\alpha }{\sqrt{n}}}[/tex]

for the combine resistance of 1458 ohms mean resistance is

[tex]\mu }=\frac{1458}{36}=40.5[/tex]

[tex]P(40.5 < x)=P(\frac{40.5-\mu }{\frac{2}{\sqrt{36}}} < \frac{x-\mu }{\frac{\alpha }{\sqrt{n}}})[/tex]

[tex]=P(\frac{40.5}{\frac{1}{3} } < Z} )[/tex]

[tex]=P(1.5 < Z)[/tex]

=[tex]1-P(1.5\leq Z)[/tex]

=1-0.9332

=0.0668

Therefore we get the probability is 0.0668.

To learn more about the probability visit:

https://brainly.com/question/24756209