Question

Suppose a certain capsule is manufactured so that the dosage of the active ingredient follows the distribution y ~ n(μ = 10 mg, σ = 1 mg). a dosage more than 13mg is considered dangerous. if we sample 49 capsules at random, compute the probability that the mean, y-bar, falls into the dangerous region. pick the closest answer be

Answer 1

So that the means of the sample falls into the dangerous region then
`{\displaystyle {\overline {Y}}}-\mu_{{\displaystyle {\overline {Y}}}} > 3`

Where
`{\displaystyle {\overline {Y}}}` is normally distributed with mean
`= \mu` and standar desviation
`\sigma_{{\displaystyle {\overline {Y}}}} =(\sigma)/(√(n))`

Then
`n = 49`

`\mu_{{\displaystyle {\overline {Y}}}}=\mu= 10`

`\sigma_{{\displaystyle {\overline {Y}}}} = (1)/(√(49) )`

So:

`P(\frac{{\displaystyle {\overline {Y}}}-\mu}{(\sigma)/(√(n))})> (3)/((1)/(√(49)))`

Finally

`P (Z> 21) = 0`.

The probability that the sample mean falls in the dangerous region is approximately zero.

This makes sense, since as the sample size increases, it is less likely to deviate more than 2 standard deviations from the mu population mean.