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QveST [7]
3 years ago
13

Suppose that 20% of adults belong to health clubs, and 51% of these health club members go to the club at least twice a week. Wh

at percent of all adults go to a health club at least twice a week? Write the information given in terms of probabilities and use the general multiplication rule. (Let H be the event that an adult belongs to a club, and T be the event that he/she goes at least twice a week. Use 3 decimal places.)
Mathematics
1 answer:
Westkost [7]3 years ago
5 0

Answer:

10.2% of adults will belong to health clubs and will go to the club at least twice a week

Step-by-step explanation:

assuming that the event H=an adult belongs to a health  club  and the event T= he/she goes at least twice a week , then if both are independent of each other:

P(T∩H)= P(H)*P(T)  ( probability of the union of independent events → multiplication rule )

replacing values

P(T∩H)= P(H)*P(T) = 0.20 * 0.51 =0.102

then  10.2% of adults will belong to health clubs and will go to the club at least twice a week

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z=1.28

And if we solve for a we got

a=3.6 +1.28*0.8=4.624

So the value of height that separates the bottom 90% of data from the top 10% is 4.624.

So then the best answer for this case would be:

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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(3.6,0.8)  

Where \mu=3.6 and \sigma=0.8

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.1   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

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