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Sveta_85 [38]
3 years ago
15

An engineer measures a sample of 1200 shafts out of a certain shipment. He finds the shafts have an average diameter of 2.45 inc

h and a standard deviation of 0.07 inch. Assume that the shaft diameter follows a Gaussian distribution. What percentage of the diameter of the total shipment of shafts will fall between 2.39inch and 2.60 inch?
Engineering
1 answer:
Vadim26 [7]3 years ago
7 0

Answer: 78.89%

Explanation:

Given : Sample size : n=  1200

Sample mean : \overline{x}=2.45

Standard deviation : \sigma=0.07

We assume that it follows Gaussian distribution (Normal distribution).

Let x be a random variable that represents the shaft diameter.

Using formula, z=\dfrac{x-\mu}{\sigma}, the z-value corresponds to 2.39 will be :-

z=\dfrac{2.39-2.45}{0.07}\approx-0.86

z-value corresponds to 2.60 will be :-

z=\dfrac{2.60-2.45}{0.07}\approx2.14

Using the standard normal table for z, we have

P-value = P(-0.86

=P(z

Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%

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A = 25 km²

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Tp = tr/2 + 0.6 tc = 1/12 + 0.96 = 1.043 hr

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Since the area is  

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0                   0                                     0

10                  4                                     1000000 m³

20                 2.5                                  625000 m³

30                 2                                      500000 m³

Total volume of runoff = 1000000 + 625000 + 500000 =  2125000 m³/s

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For the 2nd 10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

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Tb = 8/3×Tp = 8/3×1.21 = 3.227  hr

For the 3rd 10 minutes, we have

A = 25 km²

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Qp = 2.08×25×2.5/1.043 = 98.96 m³/s

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Peak flow of aggregate runoff is given by

Qp (total) = 98.96 + 123.7 +197.92 = 420.58 m³/s

Total volume of runoff is given by

Total volume of runoff = 1000000 + 625000 + 500000 =  2125000 m³/s

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