Mathematics, 16.06.2020 16:57 Christiancameron1234

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 of 13mg is considered dangerous. Compute the probability that Y falls into the dangerous region. a. 0.0013.
b. 0.013.
c. 0.9987.
d. 0.
Again suppose a certain capsule is manufactured so that the dosage of the active ingredient follows the distribution Y ∼ N(μ=10, σ=1), where units are in mg. A dosage of 13mg is considered dangerous. If we sample 49 capsules at random, compute the probability that the mean Y-bar falls into the dangerous region and pick the closest answer below. a. 0.0013.
b. 0.013.
c. 0.9987.
d. 0.

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Answer from: AsianAtari9340

(a) Probability that Y falls into the dangerous region is 0.0013.

(b) Probability that the mean Y-bar falls into the dangerous region is 0.00001.

Step-by-step explanation:

We are given that 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 of 13 mg is considered dangerous.

Let Y = dosage of the active ingredient

The z-score probability distribution for normal distribution is given by;

Z = $\frac{ Y-\mu}{\sigma} } }$ ~ N(0,1)

where, $\mu$ = population mean = 10 mg

$\sigma$ = standard deviation = 1 mg

(a) Probability that Y falls into the dangerous region is given by = P(Y $\geq$ 13 mg)

P(Y $\geq$ 13 mg) = P( $\frac{ Y-\mu}{\sigma} } }$ $\geq$ $\frac{ 13-10}{1} } }$ ) = P(Z $\geq$ 3) = 1 - P(Z < 3)

= 1 - 0.9987 = 0.0013

The above probability is calculated by looking at the value of x = 3 in the z table which has an area of 0.9987.

(b) We are given that a dosage of 13 mg is considered dangerous. And we sample 49 capsules at random.

Let $\bar Y$ = sample mean dosage

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar Y-\mu}{\frac{\sigma}{\sqrt{n} } } } }$ ~ N(0,1)

where, $\mu$ = population mean = 10 mg

$\sigma$ = standard deviation = 1 mg

n = sample of capsules = 49

So, Probability that the mean Y-bar falls into the dangerous region is given by = P( $\bar Y$ $\geq$ 13 mg)

P(Y $\geq$ 13 mg) = P( $\frac{\bar Y-\mu}{\frac{\sigma}{\sqrt{n} } } } }$ $\geq$ $\frac{13-10}{\frac{1}{\sqrt{49} } } } }$ ) = P(Z $\geq$ 21) = 1 - P(Z < 21)

= 0.00001

Answer from: Quest

b is the answer that makes most sense

step-by-step explanation:

Answer from: Quest

Multiply 9 and 67. then multiply 3 with 67. lastly the answer that you got for 9x67 is to be subtracted by the answer you got for 3x67

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