Can anyone help me to get the answers for these problems

Mathematics

1 answer:

Brut [27]3 years ago

4 0

Answer:

..........................do..................that......

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g Suppose a factory production line uses 3 machines, A, B, and C for making bolts. The total output from the line is distributed

Vladimir [108]

Answer:

The probability that it came from A, given that is defective is 0.362.

Step-by-step explanation:

Define the events:

A: The element comes from A.

B: The element comes from B.

C: The element comes from C.

D: The elemens is defective.

We are given that P(A) = 0.25, P(B) = 0.35, P(C) = 0.4. Recall that since the element comes from only one of the machines, if an element is defective, it comes either from A, B or C. Using the probability axioms, we can calculate that

$P(D) = P(A\cap D) + P(B\cap D) + P(C\cap D)$

Recall that given events E,F the conditional probability of E given F is defined as

$P(E|F) = \frac{P(E\cap F)}{P(F)}$ , from where we deduce that

$P(E\cap F) = P(E|F)P(F)$ .

We are given that given that the element is from A, the probability of being defective is 5%. That is P(D|A) =0.05. Using the previous analysis we get that

$P(D) = P(A)P(D|A)+P(B) P(D|B) + P(C)P(D|C) = 0.05\cdot 0.25+0.04\cdot 0.35+0.02\cdot 0.4 = 0.0345$

We are told to calculate P(A|D), then using the formulas we have

$P(A|D) = \frac{P(A\cap D)}{P(D)}= \frac{P(D|A)P(A)}{P(D)}= \frac{0.05\cdot 0.25}{0.0345}= 0.36231884$

3 0

3 years ago

A machine is used to fill containers with a liquid product. Fill volume can be assumed to be normally distributed. A random samp

motikmotik

Answer:

There is no statistical evidence at 1% level to accept that the mean net contents exceeds 12 oz.

Step-by-step explanation:

Given that a random sample of ten containers is selected, and the net contents (oz) are as follows: 12.03, 12.01, 12.04, 12.02, 12.05, 11.98, 11.96, 12.02, 12.05, and 11.99.

We find mean = 11.015

Sample std deviation = 3.157

a) $H_0: \bar x= 12 oz\\H_a: \bar x >12$