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

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 =

or

⇒ z₁ =

or

⇒ z₁ = - 2

similarly,

for

z-value will be

⇒ z₂ =

or

⇒ z₂ =

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%

i dont know

step-by-step explanation:

answer: procent 54 wynosi 33,33%