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%

answer: the answer is one of

step-by-step explanation:   we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following

sorry i need the points

step-by-step explanation:

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