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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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