Question

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 25 of these resistors will have a combined resistance of less than 1010 ohms?

Answer 1

The probability is
`P(\= X < 40.4) = 0.84134`

Explanation:

From the question we are told that

The mean is
`\mu =40 \ \Omega`

The standard deviation is
`\sigma = 2 \ \Omega`

The sample size is n = 25

The combined resistance is
`\sum x_i = 1010 \ \Omega`

Generally the sample mean is mathematically represented as

`\= x = (\sum x_i )/(n)`

=>
`\= x = (1010 )/(25)`

=>
`\= x = 40.4 \ \Omega`

Generally the standard error of the mean is mathematically represented as

`\sigma_(x) = (\sigma )/(√(n) )`

=>
`\sigma_(x) = ( 2 )/(√(25) )`

=>
`\sigma_(x) = 0.4`

Generally the probability that a random sample of 25 of these resistors will have a combined resistance of less than 1010 ohms is mathematically represented as

`P(\= X < 40.4) = P( (\= X - \mu )/(\sigma_(x )) < ( 40.4 - 40 )/(0.4) )`

`(\= X -\mu)/(\sigma )  =  Z (The  \ standardized \  value\  of  \ \= X )`

`P(\= X < 40.4) = P( Z < 1 )`

From the z table the area under the normal curve to the left corresponding to 1 is

`P(\= X < 40.4) = P( Z < 1 ) = 0.84134`

`P(\= X < 40.4) = 0.84134`