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A drug 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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Answer:

4.55%

Step-by-step explanation:

Data provided in the question:

Given,

Number of samples of capsules, n = 25

mean amount of ingredient, ΞΌ = 150 mg

Standard deviation, Οƒ = 5

Now,

[tex]\bar{x}_1=148\ mg[/tex]

z-value will be

β‡’ z₁ = [tex]\frac{\bar{x}_1-\mu}{(\frac{\sigma}{\sqrt n})}[/tex]

or

β‡’ z₁ = [tex]\frac{148-150}{(\frac{5}{\sqrt{25}})}[/tex]

or

β‡’ z₁ = - 2

similarly,

for [tex]\bar{x}_2=152\ mg[/tex]

z-value will be

β‡’ zβ‚‚ = [tex]\frac{\bar{x}_2-\mu}{(\frac{\sigma}{\sqrt n})}[/tex]

or

β‡’ zβ‚‚ = [tex]\frac{152-150}{(\frac{5}{\sqrt{25}})}[/tex]

or

β‡’ zβ‚‚ = 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

β‡’ 0.0455 Β or

β‡’ 0.0455 Γ— 100%

= 4.55%

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