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PilotLPTM [1.2K]
3 years ago
6

Out of 1100 discs tested 13 are defective. Estimate the number of defective discs in a batch of 41000

Mathematics
1 answer:
zvonat [6]3 years ago
5 0
I’d say 485 discs are defective out of 41000
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1) There are about 20 million people in New York State, with a total land area of about 47,000 square miles.
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A random sample of 20 students yielded a mean of ¯x = 72 and a variance of s2 = 16 for scores on a college placement test in mat
Helen [10]

Answer:

The 98% confidence interval for the variance in the pounds of impurities would be 8.400 \leq \sigma^2 \leq 39.827.

Step-by-step explanation:

1) Data given and notation

s^2 =16 represent the sample variance

s=4 represent the sample standard deviation

\bar x represent the sample mean

n=20 the sample size

Confidence=98% or 0.98

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=20-1=19

Since the Confidence is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.01,19)" "=CHISQ.INV(0.99,19)". so for this case the critical values are:

\chi^2_{\alpha/2}=36.191

\chi^2_{1- \alpha/2}=7.633

And replacing into the formula for the interval we got:

\frac{(19)(16)}{36.191} \leq \sigma^2 \leq \frac{(19)(16)}{7.633}

8.400 \leq \sigma^2 \leq 39.827

So the 98% confidence interval for the variance in the pounds of impurities would be 8.400 \leq \sigma^2 \leq 39.827.

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3 years ago
The two sides of a triangular park enclosing the right angle
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Step-by-step explanation:

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Answer:

A) WTUV is moved onto W'T'U'V' after translating −15 units vertically, and then rotating 180° counterclockwise around the origin.

Step-by-step explanation:

WTUV is moved onto W'T'U'V' after translating −15 units vertically, and then rotating 180° counterclockwise around the origin.

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