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ELEN [110]
3 years ago
15

A production process manufactures electronic components with timing signals whose duration follows a normal distribution. A rand

om sample of 55 components was​ taken, and the durations of their timing signals were measured.a. The probability is 0.01 that the sample variance is bigger than what percentage of the population​ variance?
b. The probability is 0.05 that the sample variance is less than what percentage of the population​ variance?
Mathematics
1 answer:
REY [17]3 years ago
8 0

Answer:

Check the explanation

Step-by-step explanation:

b ) P(s^2

or , P\left (\frac{(n-1)s^2}{\sigma^2}

or , P\left (\chi^2_5

or , 5p =1.145476

or , p =\frac{1.145476}{5}= 0.2290952

Required percentage = 22.91

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5.5%

Step-by-step explanation:

To solve this problem we can use a modified version of the simple interest formula which is shown below:

r=\frac{I}{Pt}

<em>I = interest amount</em>

<em>P = principal amount</em>

<em>t = time (years)</em>

<em />

The first step is to find the interest gained from the investment.

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Next, plug in the values into the equation:

r=\frac{673.75}{(2,450)(5)}     Multiply the bottom values

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The last step is to convert 0.055 into a percent:

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