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

Step 1
Standard normal distribution:
The typical normal distribution is a specific instance of the 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:
$z=\frac{y-E(y)}{S.D(y)}=\frac{y-\mu }{s}-N(0.1)$
Step 2
What is the probability that the current measurement exceeds 13 milliamperes:
Consider the current measurements in a strip of wire be denoted by y. It is assumed 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 required current in a wire should be more than 13 milliamperes. That is, y > 13.
The value of $P(y>13)$ is obtained as 0.06681 from the calculation given below:
$P(y>13)=1-P(y\le 13)$
$=1-P((y-\frac{10}{2})\le \frac{13-10}{2})$
$=1-P((y-E\frac{y}{S}.D(y))\le (1.5)$
$=1-P(Z\le 1.5)$
$\cong 1-\varphi (1.5),[Fromcthetablevaluesofstandardnormaldistribution.Theareacorrespondingtoleftof1.5is0.06681]$
$\cong 1-0.93319\cong 0.06681$
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, $9<y<11$.
The value of $P(9<y<11)$ is obtained as 0.38292 from the calculation given below:
$P(9<y<11)=P(\frac{9-10}{2}<(y-\frac{10}{2})<\frac{11-10}{2})$
$=P(-0.5<(y-E\frac{y}{S.D(y)}<(0.5)$
$=P(-0.5<Z<0.5)$
$=\varphi (0.5)-\varphi (-0.5)$
$\cong 2\varphi (0.5)-1$
$\cong (2\times 0.69146)-1,[Fromthetablevaluesofstandardnormaldistribution.Theareacorrespondingtoleftof0.5is0.69146]$
$\cong 0.38292$
Step 4
Answer:
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.

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

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

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 $\le$ 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