Question #60a21

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Question #60a21

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Answer 1 · 2016-05-20T03:48:01

Probability that 100 or more out of the 400 will break $= 0.0062$ $=0.0062$

Explanation:

First find the Mean

$¯ x = n p = 400 \times 0.2 = 80$ $barx=np=400xx0.2=80$

$q = 1 - p = 1 - 0.2 = 0.8$ $q=1-p=1-0.2=0.8$

Find the Standard Deviation. In our case it is the Standard Error.

$S E = \sqrt{n p q} = \sqrt{400 \times 0.2 \times 0.8} = \sqrt{64} = 8$ $SE=sqrt(npq)=sqrt(400xx0.2xx0.8)=sqrt64=8$

Calculate the Standard Error

$S E = \frac{x - ¯ x}{S E} = \frac{100 - 80}{8} = \frac{20}{8} = 2.5$ $SE=(x-barx)/(SE)=(100-80)/8=20/8=2.5$

Look at the diagram.

The green shaded area represents the probability that 100 or more out of the 400 will break.

Refer the 'Area under Normal Curve'. Find the value for 2.5. It is 0.4938. The blue shaded area represents the probability value for $x = 80$ $x=80$ to $x = 100$ $x=100$

The right wing of the curve represent probability value $0.5$ $0.5$

The probability that 100 or more out of the 400 will break $= 0.5 - 0.4938 = 0.0062$ $=0.5-0.4938=0.0062$

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Question #60a21

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