Physics, asked by sreedarbabut2016, 1 month ago

The guaranteed average life of a certain type of electric light bulb is 1000 hrs with a standard deviation of 125hrs. It is decided to sample the output so as to ensure that 90% of the bulbs do not fall short of the guaranteed average by more than 2.5% find the minimum sample size.​

Answers

Answered by mevadajeet5999
2

Answer:

I do not know the answer ok

Answered by mad210201
6

Given:

guaranteed average life \mu=1000\ hrs

standard deviation \sigma=125\ hrs

As we do not want the sample mean to be less than the guaranteed average mean by more than 2.5 %

To Find:

minimum sample size

Solution:

We know that

\bar x>1000 - 2.5 % of 1000\\\\\Rightarrow \bar x>1000 - 25=975

Assume n be the given sample size.

Then,

Z=\dfrac{\bar x-\mu}{\sigma/\sqrt{n} }\sim N(0,1),

The sample is large

Z=\dfrac{\bar x-\mu}{\sigma/\sqrt{n} }>\dfrac{975-1000}{125/\sqrt{n} }\\=-\dfrac{\sqrt{n} }{5}

Use the given condition:

P(Z>-\frac{\sqrt{n} }{5})=0.90\\\Rightarrow P(0<Z<\frac{\sqrt{n} }{5})=0.90

\Rightarrow\dfrac{\sqrt{n} }{5}=0.90\\n=25\times(0.90)^{2}\\n=25\times0.81\\n=25.

The minimum sample size is 25.

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