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New York City is known for it's tourist attractions and high priced real estate. The mean hotel room rate is $202 per night. Assume that the room rates are normally distributed with a standard deviation of $70.What is the probability that a hotel room costs between $210 and $290?

User MarkPlewis
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1 Answer

2 votes

Answer:


P(210<X<290)=P((210-\mu)/(\sigma)<(X-\mu)/(\sigma)<(290-\mu)/(\sigma))=P((210-202)/(70)<Z<(290-202)/(70))=P(0.114<z<1.26)

And we can find this probability with this difference


P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)

And we can find the difference with the normal standard distirbution or excel:


P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)=0.896-0.545=0.351

Explanation:

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


X \sim N(202,70)

Where
\mu=202 and
\sigma=70

We are interested on this probability


P(210<X<290)

The z score formula is given by:


z=(x-\mu)/(\sigma)

Using the formula we got:


P(210<X<290)=P((210-\mu)/(\sigma)<(X-\mu)/(\sigma)<(290-\mu)/(\sigma))=P((210-202)/(70)<Z<(290-202)/(70))=P(0.114<z<1.26)

And we can find this probability with this difference


P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)

And we can find the difference with the normal standard distirbution or excel:


P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)=0.896-0.545=0.351

User Alon Or
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