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ahrozycki
02.11.2019 •
Mathematics
Adrug company produces capsules that are designed to contain a specified amount of the active ingredient. the mean amount of the ingredient in one capsule is 150 mg, the sd is 5 mg and the distribution of the amount is normal. a sample of 25 capsules is selected at random, and the production process is stopped if the average amount of the active ingredient in this sample is outside of the range [148 mg to 152 mg]. what is the probability (approximately) that the production is stopped
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Ответ:
4.55%
Step-by-step explanation:
Data provided in the question:
Given,
Number of samples of capsules, n = 25
mean amount of ingredient, μ = 150 mg
Standard deviation, σ = 5
Now,
z-value will be
⇒ z₁ =![\frac{\bar{x}_1-\mu}{(\frac{\sigma}{\sqrt n})}](/tpl/images/0356/4945/4f85b.png)
or
⇒ z₁ =![\frac{148-150}{(\frac{5}{\sqrt{25}})}](/tpl/images/0356/4945/d5a52.png)
or
⇒ z₁ = - 2
similarly,
for![\bar{x}_2=152\ mg](/tpl/images/0356/4945/2e8be.png)
z-value will be
⇒ z₂ =![\frac{\bar{x}_2-\mu}{(\frac{\sigma}{\sqrt n})}](/tpl/images/0356/4945/802a4.png)
or
⇒ z₂ =![\frac{152-150}{(\frac{5}{\sqrt{25}})}](/tpl/images/0356/4945/ad80c.png)
or
⇒ z₂ = 2
Now,
P( -2 < x < 2) = P( z < 2) - P(z < -2)
from the z-value vs P table, we have
= 0.9772498 - 0.0227501
= 0.9545
therefore,
Probability that the production is stopped = 1 - 0.9545
⇒ 0.0455 or
⇒ 0.0455 × 100%
= 4.55%
Ответ:
what?
Step-by-step explanation:
where is the freakin' question?