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3 years ago

5

A factory makes tennis balls. Every tennis ball must meet a quality standard, or it will be rejected. The percentage that meets

the quality standard is 99.8%. If 20,000 tennis balls are made, then about how many are expected to be rejected?

1 answer:

Anettt [7]3 years ago

3 0

The answer you're looking for is 40 tennis balls.
Hope this helped!

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3 years ago

Consider a system with one component that is subject to failure, and suppose that we have 115 copies of the component. Suppose f

castortr0y [4]

Answer:

Step-by-step explanation:

From the given information:

the mean $(\mu) = 115 \times 20$

= 2300

Standard deviation = $20 \times \sqrt{115}$

Standard deviation (SD) = 214.4761

TO find:

a) $P(x > 3500)= P(Z > \dfrac{3500-\mu}{214.4761})$

$P(x > 3500)= P(Z > \dfrac{3500-2300}{214.4761})$

$P(x > 3500)= P(Z > \dfrac{1200}{214.4761})$

$P(x > 3500)= P(Z >5.595)$

From the Z-table, since 5.595 is > 3.999

$P(x > 3500)=1-0.9999$

P(x > 3500) = 0.0001

b)

Here, the replacement time for the mean $(\mu) = \dfrac{0+0.5}{2}$

= 0.25

Replacement time for the Standard deviation $\sigma = \dfrac{0.5-0}{\sqrt{12}}$

$\sigma = 0.1443$

For 115 component, the mean time = (115 × 20)+(114×0.25)

= 2300 + 28.5

= 2328.5

Standard deviation = $\sqrt{(115\times 20^2) +(114\times (0.1443)^2)}$

= $\sqrt{(115\times 400) +(114\times 0.02082249}$

= $\sqrt{(46000) +2.37376386}$

= $\sqrt{(46000) +(2.37376386)}$

= $\sqrt{46002.374}$

= 214.482

Now; the required probability:

$P(x > 4125) = P(Z > \dfrac{4125- 2328.5}{214.482})$

$P(x > 4125) = P(Z > \dfrac{1796.5}{214.482})$

$P(x > 4125) = P(Z >8.376)$

$P(x > 4125) =1- P(Z$

From the Z-table, since 8.376 is > 3.999

P(x > 4125) = 1 - 0.9999

P(x > 4125) = 0.0001

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