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pogonyaev
15 days ago
11

The mean diameter of the rim of Honda tires is 16 inches. Assume that the standard deviation of diameter of the rims is 0.3 inch

es. For quality control purposes, the diameter of the rims of 9 tires is measured every hour. The manager applies the rule that if the mean of diameter of a rim is greater or equal to 16.25, and lesser or equal to 15.75, the manufacturing should be stopped. If the diameter is between 15.75 and 16.25, the manufacturing process is not to be disturbed. a. Calculate the probability of stopping the manufacturing when the sample mean is 16 inches. b. Calculate the probability of stopping the manufacturing in case the mean is shifted to 16.05 inches. c. Calculate the probability of not disturbing the manufacturing if mean shifts to 16.25 inches.
Mathematics
1 answer:
Leona [12.6K]15 days ago
3 0

Sagot:

a. 0.0124 = 1.24% ang posibilidad na itigil ang paggawa kapag ang sample mean ay 16 inches.

b. 0.0241 = 2.41% ang posibilidad na itigil ang paggawa kung ang mean ay lumipat sa 16.05 inches.

c. 0.5 = 50% ang posibilidad na hindi makagambala sa produksyon kung ang mean ay lumipat sa 16.25 inches.

Hakbang-hakbang na paliwanag:

Upang masagot ang tanong na ito, kinakailangan nating maunawaan ang normal probability distribution at ang central limit theorem.

Normal probability distribution

Kapag ang distribusyon ay normal, ginagamit natin ang z-score formula.

Sa isang set na may mean \mu at standard deviation \sigma, ang z-score ng sukat X ay ibinibigay ng:

Z = \frac{X - \mu}{\sigma}

Ang Z-score ay sumusukat kung gaano karaming standard deviations ang sukat ay mula sa mean. Pagkatapos makuha ang Z-score, titingnan natin ang z-score table at hahanapin ang p-value na kaugnay ng z-score na ito. Ang p-value na ito ay ang posibilidad na ang halaga ng sukat ay mas maliit kaysa sa X, na ibig sabihin ay ang percentile ng X. Sa pamamagitan ng pagbawas ng 1 sa p-value, makakakuha tayo ng posibilidad na ang halaga ng sukat ay mas mataas sa X.

Central Limit Theorem

Itinataguyod ng Central Limit Theorem na, para sa isang normal na ipinamamahaging random variable X, na may mean \mu at standard deviation \sigma, ang sampling distribution ng sample means na may sukat n ay maaaring i-approximate sa isang normal na distribusyon na may mean \mu at standard deviation s = \frac{\sigma}{\sqrt{n}}.

Ipagpalagay na ang standard deviation ng diameter ng mga rims ay 0.3 inches. Ang samples ay 9.

Ibig sabihin nito \sigma = 0.3, n = 9, s = \frac{0.3}{\sqrt{9}} = 0.1

a. Kalkulahin ang posibilidad na itigil ang paggawa kapag ang sample mean ay 16 inches.

\mu = 16Narito ang

Mas mataas sa 16.25:

Ito ay 1 na ibinawas mula sa p-value ng Z kapag X = 16.25. Kaya

Z = \frac{X - \mu}{\sigma}

Sa pamamagitan ng Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{16.25 - 16}{0.1}

Z = 2.5

Z = 2.5may p-value na 0.9938

1 - 0.9938 = 0.0062

Mas mababa sa 15.75:

Ito ay p-value ng Z kapag X = 15.75. Kaya

Z = \frac{X - \mu}{s}

Z = \frac{15.75 - 16}{0.1}

Z = -2.5

Z = -2.5may p-value na 0.0062

Probabilidad ng pagtigil:

2*0.0062 = 0.0124

0.0124 = 1.24% ang posibilidad na itigil ang paggawa kapag ang sample mean ay 16 inches.

b. Kalkulahin ang posibilidad na itigil ang paggawa kung ang mean ay lumipat sa 16.05 inches.

\mu = 16.05

Narito ang

Mas mataas sa 16.25:

Z = \frac{X - \mu}{s}

Z = \frac{16.25 - 16.05}{0.1}

Z = 2

may p-value na 0.9772

1 - 0.9772 = 0.0228

Mas mababa sa 15.75:

Ito ang p-value ng Z kapag X = 15.75. Kumbaga

Z = \frac{X - \mu}{s}may p-value na 0.0013

Z = \frac{15.75 - 16.05}{0.1}

Z = -3

Z = -3Probabilidad ng pagtigil:

0.0228 + 0.0013 = 0.0241

0.0241 = 2.41% ang posibilidad na itigil ang paggawa kung ang mean ay lumipat sa 16.05 inches.

c. Kalkulahin ang posibilidad na hindi makagambala sa produksyon kung ang mean ay lumipat sa 16.25 inches.

Sa pagitan ng 16.25 at 15.75 na may \mu = 16.25. Ito ang p-value ng Z kapag X = 16.25 na ibinawas mula sa p-value ng Z kapag X = 15.75.

X = 16.25

Z = \frac{X - \mu}{s}

Z = \frac{16.25 - 16.25}{0.1}

Z = 0

Z = 0may p-value na 0.5

X = 15.75

Z = \frac{X - \mu}{s}

Z = \frac{15.75 - 16.25}{0.1}

Z = -5

Z = -5may p-value na 0

0.5 - 0 = 0.5

0.5 = 50% ang posibilidad na hindi makagambala sa produksyon kung ang mean ay lumipat sa 16.25 inches.

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