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

Burhan Hopper 2021-02-14 Answered
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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Expert Answer

Demi-Leigh Barrera
Answered 2021-02-15 Author has 97 answers

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=yE(y)S.D(y)=yμsN(0.1)
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(y>13) is obtained as 0.06681 from the calculation given below:
P(y>13)=1P(y13)
=1P((y102)13102)
=1P((yEyS.D(y))(1.5)
=1P(Z1.5)
1ϕ(1.5),[Fromcthe table values of standard normal distribution. The area corresponding to left of 1.5 is 0.06681]
10.933190.06681
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(9102<(y102)<11102)
=P(0.5<(yEyS.D(y)<(0.5)
=P(0.5<Z<0.5)
=ϕ(0.5)ϕ(0.5)
2ϕ(0.5)1
(2×0.69146)1,[From the table values of standard normal distribution. The area corresponding to left of 0.5 is 0.69146]
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.

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Jeffrey Jordon
Answered 2021-10-08 Author has 2087 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=(xμ)σ

Where

x = current measurements in a strip.

μ= mean current

σ = standard deviation

From the information given,

μ=10

σ=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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