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maw [93]
3 years ago
9

The weight of a product is normally distributed with a mean of four ounces and a variance of .25 squared ounces. What is the pro

bability that a randomly selected unit from a recently manufactured batch weighs more than 3.75 ounces
Mathematics
1 answer:
ololo11 [35]3 years ago
5 0

Answer:

P(X>3.75)=P(\frac{X-\mu}{\sigma}>\frac{3.75-\mu}{\sigma})=P(Z>\frac{3.75-4}{0.5})=P(z>-0.5)

And we can find this probability using the complement rule and we got:

P(z>-0.5)=1-P(z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(4,\sqrt{0.25}=0.5)  

Where \mu=4 and \sigma=0.5

We are interested on this probability

P(X>3.75)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>3.75)=P(\frac{X-\mu}{\sigma}>\frac{3.75-\mu}{\sigma})=P(Z>\frac{3.75-4}{0.5})=P(z>-0.5)

And we can find this probability using the complement rule and we got:

P(z>-0.5)=1-P(z

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(a) A_1 and A_2 are indeed mutually-exclusive.

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Step-by-step explanation:

<h3>(a)</h3>

P(A_1 \; \cap \; A_2) = 0 means that it is impossible for events A_1 and A_2 to happen at the same time. Therefore, event A_1 and A_2 are mutually-exclusive.

<h3>(b)</h3>

By the definition of conditional probability:

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<h3>(c)</h3>

Note that:

\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}.

In other words, A_1 and A_2 are collectively-exhaustive. Since A_1 and A_2 are collectively-exhaustive and mutually-exclusive at the same time:

\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}.

<h3>(d)</h3>

By Bayes' Theorem:

\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}.

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