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Lera25 [3.4K]
3 years ago
10

The average time a subscriber spends reading The Wall Street Journal is 49 minutes (The Wall Street Journal Subscriber Study). A

ssume the standard deviation is 16 minutes and that the times are normal are normally distribution. 1) What is the probability a subscriber will spend at least 1 hour reading the journal?
Mathematics
1 answer:
erica [24]3 years ago
5 0

Answer:

P(X>60)=P(\frac{X-\mu}{\sigma}>\frac{60-\mu}{\sigma})=P(Z>\frac{60-49}{16})=P(z>0.6875)

And we can find the probability with the complement rule and the normal standard distirbution and we got:

P(z>0.6875)=1-P(z

Step-by-step explanation:

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

X \sim N(49,16)  

Where \mu=49 and \sigma=16

We are interested on this probability

P(X>60)

And we can use the z score formula given by:

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

Using this formula we got:

P(X>60)=P(\frac{X-\mu}{\sigma}>\frac{60-\mu}{\sigma})=P(Z>\frac{60-49}{16})=P(z>0.6875)

And we can find the probability with the complement rule and the normal standard distirbution and we got:

P(z>0.6875)=1-P(z

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F(x)=-x^2+8x+15 ...?
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Read 2 more answers
Some help please stuck on 4, 8, 9
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4:
a) if a cloud is a mid-level cloud, then it is between 6,500 and 20,000 feet.
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8:
P=2l+2w
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Add all x terms together, and the numbers together
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Now, plug in this x value into the formulas for the two angles
ABD = 2(22) + 10 = 44+10 = 54 degrees
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