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%

not enough information

step-by-step explanation:

we cannot know this, because the other students could have handed out 10 flyers, or they could have handed out 100. one way, the mean would go up, and another way, the mean would go down.

a

step-by-step explanation: