# Suppose that the current measurements in a strip of wire are assumed to follow a normal distribution with a mean of 10 milliamperes and a standard dev

Suppose that the current measurements in a strip of wire are assumed to follow a normal distribution with a mean of 10 milliamperes and a standard deviation of 2 milliamperes.
What is the probability that a measurement exceeds 13 milliamperes? What is the probability that a current measurement is between 9 and 11 milliamperes.
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Demi-Leigh Barrera

Step 1
Standard normal distribution:
The standard normal distribution is a special case of normal distribution. The standard normal distribution will have mean 0 and standard deviation 1. If a random variable y follows normal distribution with mean (m) and standard deviation (s), then the standard normal variable z will be as given below:
$z=\frac{y-E\left(y\right)}{S.D\left(y\right)}=\frac{y-\mu }{s}-N\left(0.1\right)$
Step 2
Find the probability that the current measurement exceeds 13 milliamperes:
Consider the current measurements in a strip of wire be denoted by y. It is given that the current measurements in a strip of wire are normally distributed.
The mean and standard deviation for current measurements in a strip of wire is y-bar = 10 milliamperes and s(y) = 2 milliamperes, respectively.
The given requirement is that the current measurement in a strip of wire should be greater than 13 milliamperes. That is, y > 13.
The value of $P\left(y>13\right)$ is obtained as 0.06681 from the calculation given below:
$P\left(y>13\right)=1-P\left(y\le 13\right)$
$=1-P\left(\left(y-\frac{10}{2}\right)\le \frac{13-10}{2}\right)$
$=1-P\left(\left(y-E\frac{y}{S}.D\left(y\right)\right)\le \left(1.5\right)$
$=1-P\left(Z\le 1.5\right)$

$\cong 1-0.93319\cong 0.06681$
Step 3
Find the probability that the current measurement lies between 9 and 11 milliamperes:
The given requirement is that the current measurement in a strip of wire should lie between 9 and 11 milliamperes. That is, $9.
The value of $P\left(9 is obtained as 0.38292 from the calculation given below:
$P\left(9
$=P\left(-0.5<\left(y-E\frac{y}{S.D\left(y\right)}<\left(0.5\right)$
$=P\left(-0.5
$=\varphi \left(0.5\right)-\varphi \left(-0.5\right)$
$\cong 2\varphi \left(0.5\right)-1$

$\cong 0.38292$
Step 4
The probability that the current measurement exceeds 13 milliamperes is 0.06681.
The probability that the current measurement lies between 9 and 11 milliamperes is 0.38292.

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Jeffrey Jordon

Suppose that the current measurements in a strip of wire are assumed to follow a normal distribution, we would apply the formula for normal distribution which is expressed as

$z=\frac{\left(x-\mu \right)}{\sigma }$

Where

x = current measurements in a strip.

$\mu$= mean current

$\sigma$ = standard deviation

From the information given,

$\mu =10$

$\sigma =2$

We want to find the probability that a measurement exceeds 13 milliamperes. It is expressed as

P(x > 13) = 1 - P(x \leq 13)

For x = 13,

z = (13 - 10)/2 = 1.5

Looking at the normal distribution table, the probability corresponding to the z score is 0.933

P(x > 13) = 1 - 0.933 = 0.067

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