Burhan Hopper

2021-02-14

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.

What is the probability that a measurement exceeds 13 milliamperes? What is the probability that a current measurement is between 9 and 11 milliamperes.

Demi-Leigh Barrera

Skilled2021-02-15Added 97 answers

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:

$P(y>13)=1-P(y\le 13)$

$=1-P((y-\frac{10}{2})\le \frac{13-10}{2})$

$=1-P((y-E\frac{y}{S}.D(y))\le (1.5)$

$=1-P(Z\le 1.5)$

$\cong 1-\varphi (1.5),[Fromcthe\text{}table\text{}values\text{}of\text{}standard\text{}normal\text{}distribution.\text{}The\text{}area\text{}corresponding\text{}to\text{}left\text{}of\text{}1.5\text{}is\text{}0.06681]$

$\cong 1-0.93319\cong 0.06681$

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:

$P(9<y<11)=P(\frac{9-10}{2}<(y-\frac{10}{2})<\frac{11-10}{2})$

$=P(-0.5<(y-E\frac{y}{S.D(y)}<(0.5)$

$=P(-0.5<Z<0.5)$

$=\varphi (0.5)-\varphi (-0.5)$

$\cong 2\varphi (0.5)-1$

$\cong (2\times 0.69146)-1,[From\text{}the\text{}table\text{}values\text{}of\text{}standard\text{}normal\text{}distribution.\text{}The\text{}area\text{}corresponding\text{}to\text{}left\text{}of\text{}0.5\text{}is\text{}0.69146]$

$\cong 0.38292$

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.

Jeffrey Jordon

Expert2021-10-08Added 2605 answers

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

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