If a Variable has a normal distribution with mean 30 and standard deviation 5 find the probability that the variable will be between 25 and 37.

A. 0.77
B. 80
0.72
0.75

Answers

Answer 1

Answer:

P(25 < x < 37) = 0.77

Step-by-step explanation:

Given - If a Variable has a normal distribution with mean 30 and standard deviation 5

To find - find the probability that the variable will be between 25 and 37.

Proof -

Given that,

Mean, μ = 30

S.D, σ = 5

Now,

[tex]z = \frac{x-\mu}{\sigma}[/tex] ~ N(0,1)

Now,

P(25 < x < 37)

= [tex]P(\frac{25 - 30}{5} < z < \frac{37 - 30}{5} )[/tex]

= P(1 < z < 1.4)

= P(z < 1.4) - P(z < -1)

= 0.9192 - 0.1587

= 0.7605

≈ 0.77

∴ we get

P(25 < x < 37) = 0.77


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Step-by-step explanation:

Given Function:

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Step-by-step explanation:

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Step-by-step explanation:

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First we must identify the relationship between weight of gold and cost of gold.

We can do this by dividing cost by corresponding weight.

7200 / 12 = 600

13200 / 22 = 600

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each proportion has an outcome of 600

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Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

Sample size, n = 20

Sample _ tread wear, x __ Range

1 ______ 33.67 ___ 14

2______ 26.33 ___ 17

3 _____ 29.67 ____ 9

4 ______ 21 ______ 8

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The R chart control limit is given by :

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From the R chart :

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Answers

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Answers

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Answer:

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