A the length of similar components produced by a company are approximately by a normal distribution model with a mean of 5 cm standard deviation of 0.02cm IF a components is chosen at random. a) what is the probability that the lenght of this component 4.98 and 5.02cm.(b) what is the probability that the lenght of this component is between 4.96 and 5.04cm
Answers
Answer:
(a) ANSWER
Given,
μ=5(mean)
σ=0.02 (standard deviation)
find the probability for 4.98<x<5.02
converting the problem in standard form
Z=(x−μ)/σ
for x=4.98,
Z=−1
for x=5.02,
Z=1
for finding the probability for 4.98<x<5.02
in the standard form −1<z<1
in standard form mean is equal to zero and the standard deviation is 1.
ao we will have find the probability for (μ−σ) to(μ+σ)
i.e. −1 to+1
and probability for (μ−σ) to(μ+σ) is 0.6826
(b)ANSWER


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Class 12
>>Applied Mathematics
>>Standard Probability Distributions
>>Normal Distribution
>>The length of similar components produce
Question

The length of similar components produced by a company is approximated by a normal distribution model with a mean of 5 cm and a standard deviation of 0.02 cm. If a component is chosen at random, what is the probability that the length of this component is between 4.96 and 5.04 cm?
A
0.9544
B
0.1236
C
0.7265
D
0.9546
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Solution

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Correct option is A)
Let x be the length of the component.
x has μ=5,σ=0.02
We need to find the probability of the length of the component between 4.96 and 5.04. That is to find P(4.96<x<5.04).
Given x,z=σx−μ
Thus for x=4.96,z=0.024.96−5=−2
and for x=5.04,z=0.025.04−5=2
Therefore P(4.96<x<5.04)=P(−2<z<2)
=P(z<2)−P(z<−2)
=0.9772−0.0228 (from normal distribution table)
=0.9544
∴P(4.96<x<5.04)=0.9544
Hence the probability of the length of the component between
Explanation:
THANK YOU
Answer:
Correct option is A)
Let x be the length of the component.
x has μ=5,σ=0.02
We need to find the probability of the length of the component between 4.96 and 5.04. That is to find P(4.96<x<5.04).
Given x,z=
σ
x−μ
Thus for x=4.96,z=
0.02
4.96−5
=−2
and for x=5.04,z=
0.02
5.04−5
=2
Therefore P(4.96<x<5.04)=P(−2<z<2)
=P(z<2)−P(z<−2)
=0.9772−0.0228 (from normal distribution table)
=0.9544
∴P(4.96<x<5.04)=0.9544
Hence the probability of the length of the component between 4.96 and 5.04 is 0.9544
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