Respuesta :
Answer:
(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 = [tex]\frac{ Y-\mu}{\sigma} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean = 10 mg
[tex]\sigma[/tex] = standard deviation = 1 mg
(a) Probability that Y falls into the dangerous region is given by = P(Y [tex]\geq[/tex] 13 mg)
P(Y [tex]\geq[/tex] 13 mg) = P( [tex]\frac{ Y-\mu}{\sigma} } }[/tex] [tex]\geq[/tex] [tex]\frac{ 13-10}{1} } }[/tex] ) = P(Z [tex]\geq[/tex] 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 [tex]\bar Y[/tex] = sample mean dosage
The z-score probability distribution for sample mean is given by;
Z = [tex]\frac{\bar Y-\mu}{\frac{\sigma}{\sqrt{n} } } } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean = 10 mg
[tex]\sigma[/tex] = 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([tex]\bar Y[/tex] [tex]\geq[/tex] 13 mg)
P(Y [tex]\geq[/tex] 13 mg) = P( [tex]\frac{\bar Y-\mu}{\frac{\sigma}{\sqrt{n} } } } }[/tex] [tex]\geq[/tex] [tex]\frac{13-10}{\frac{1}{\sqrt{49} } } } }[/tex] ) = P(Z [tex]\geq[/tex] 21) = 1 - P(Z < 21)
= 0.00001
