The lengths of text messages are normally distributed with an unknown population mean. A random sample of text messages is taken
2 answers:
Answer:
95% of the text messages have length between 23 units and 47 units.
Step-by-step explanation:
We are given the following in the question:
The lengths of text messages are normally distributed.
95% confidence interval:
(23,47)
Thus, we could interpret the confidence interval as:
About 95% of the text messages have length between 23 units and 47 units.
By Empirical rule for a normally distributed data, about 95% of data lies within 2 standard deviations of mean , thus we can write:
Thus, the mean length of text messages is 23 units and standard deviation is 6 units.
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The volume of a sphere is given by:
So, we need to deduct this equation. We will walk through Calculus on the concept of a solid of revolution that is a solid figure that is obtained by rotating a plane curve around some straight line (the axis of revolution<span>) that lies on the same plane. We know from calculus that:
</span>![V=\pi \int_{a}^{b}[f(x)]^{2}dx](https://tex.z-dn.net/?f=V%3D%5Cpi%20%5Cint_%7Ba%7D%5E%7Bb%7D%5Bf%28x%29%5D%5E%7B2%7Ddx)
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Then, according to the concept of solid of revolution we are going to rotate a circumference shown in the figure, then:
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</span>
<span>
Then, according to the concept of solid of revolution we are going to rotate a circumference shown in the figure, then:
</span>
<span>
Isolationg y:
</span>
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<span>
</span>
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</span>
<span>
Solving:
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Check the tree diagram, of all the possible scenarios and the probabilities.
All the possible scenarios are:
1st.non defective, 2nd.non defective
1st.non defective, 2nd. defective
1st. defective, 2nd. non defective
1st. defective, 2nd. defective
consider the case 1st.non defective, 2nd.non defective
the probability that the first one is non defective is 8/10, since 8 out of 10 are non defective.
In this scenario, the second one is non defective as well. The probability for this to happen is 7/9, since now we have 7 non defective pots, and 9 in total.
This means that the probability of the first to be non defective, and the second non-defective is (8/10) * (7/9)=0.622
similarly we calculate the other cases, as shown in the picture.
P(at least one is defective)=
P(1st.non defective, 2nd. defective)+P(1st. defective, 2nd. non defective)+P(1st. defective, 2nd. defective)=
0.178+0.178+0.022=0.378
Answer: 0.378
All the possible scenarios are:
1st.non defective, 2nd.non defective
1st.non defective, 2nd. defective
1st. defective, 2nd. non defective
1st. defective, 2nd. defective
consider the case 1st.non defective, 2nd.non defective
the probability that the first one is non defective is 8/10, since 8 out of 10 are non defective.
In this scenario, the second one is non defective as well. The probability for this to happen is 7/9, since now we have 7 non defective pots, and 9 in total.
This means that the probability of the first to be non defective, and the second non-defective is (8/10) * (7/9)=0.622
similarly we calculate the other cases, as shown in the picture.
P(at least one is defective)=
P(1st.non defective, 2nd. defective)+P(1st. defective, 2nd. non defective)+P(1st. defective, 2nd. defective)=
0.178+0.178+0.022=0.378
Answer: 0.378