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3 years ago

5

Five percent of canned goods get bumps or dents in the cans during manufacturing. Let X = the number of cans that are inspected

before one with a bump or dent is found.
What is the probability that more than 3 cans are inspected before one with a bump or dent is found?

Round to 4 decimal places.

1 answer:

Ainat [17]3 years ago

5 0

Answer:

0.8574

Step-by-step explanation:

I got it correct1

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Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true

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Answer:

(a) 95% confidence interval for the true average porosity of a certain seam is [4.52 , 5.18].

(b) 98% confidence interval for the true average porosity of a another seam is [4.12 , 4.99].

Step-by-step explanation:

We are given that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation 0.75.

(a) Also, the average porosity for 20 specimens from the seam was 4.85.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average porosity = 4.85

$\sigma$ = population standard deviation = 0.75

n = sample of specimens = 20

$\mu$ = true average porosity

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.</em>

<u>So, 95% confidence interval for the true mean, </u> $\mu$ <u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

<u>95% confidence interval for</u> $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $4.85-1.96 \times {\frac{0.75}{\sqrt{20} } }$ , $4.85+1.96 \times {\frac{0.75}{\sqrt{20} } }$ ]

= [4.52 , 5.18]

Therefore, 95% confidence interval for the true average porosity of a certain seam is [4.52 , 5.18].

(b) Now, there is another seam based on 16 specimens with a sample average porosity of 4.56.

The pivotal quantity for 98% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average porosity = 4.56

$\sigma$ = population standard deviation = 0.75

n = sample of specimens = 16

$\mu$ = true average porosity

<em>Here for constructing 98% confidence interval we have used One-sample z test statistics as we know about population standard deviation.</em>

<u>So, 98% confidence interval for the true mean, </u> $\mu$ <u> is ;</u>

P(-2.3263 < N(0,1) < 2.3263) = 0.98 {As the critical value of z at 1% level

of significance are -2.3263 & 2.3263}

P(-2.3263 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.3263$ ) = 0.98

P( $\bar X-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.98

<u>98% confidence interval for</u> $\mu$ = [ $\bar X-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $4.56-2.3263 \times {\frac{0.75}{\sqrt{16} } }$ , $4.56+2.3263 \times {\frac{0.75}{\sqrt{16} } }$ ]

= [4.12 , 4.99]

Therefore, 98% confidence interval for the true average porosity of a another seam is [4.12 , 4.99].

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3 years ago

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Answer:

$Blue =1\frac{3}{5}$

Step-by-step explanation:

Given

$Blue = 1\frac{1}{3}$

$Red = 1\frac{2}{3}$

Required

Determine the amount of blue paint when Red = 2

To do this, we make use of the following equivalent ratios

$Ratio = Blue : Red$

When $Blue = 1\frac{1}{3}$ and $Red = 1\frac{2}{3}$

$Ratio = 1\frac{1}{3} : 1\frac{2}{3}$

When Red = 2

$Ratio = Blue : 2$

Equate both ratios

$Blue : 2 = 1\frac{1}{3} : 1\frac{2}{3}$

Convert to division

$\frac{Blue}{2} =1\frac{1}{3} / 1\frac{2}{3}$

Convert fractions to improper fraction

$\frac{Blue}{2} =\frac{4}{3} / \frac{5}{3}$

$\frac{Blue}{2} =\frac{4}{3} * \frac{3}{5}$

$\frac{Blue}{2} =\frac{4}{5}$

Make Blue the subject

$Blue =\frac{4}{5}*2$

$Blue =\frac{8}{5}$

$Blue =1\frac{3}{5}$

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3 years ago