MATHEMATICS COLLEGE

Television viewing reached a new high when the global information and measurement company reported a mean daily viewing time of hours per household. Use a normal probability distribution with a standard deviation of hours to answer the following questions about daily television viewing per household. a. What is the probability that a household views television between 3 and 9 hours a day (to 4 decimals)? b. How many hours of television viewing must a household have in order to be in the 2%top of all television viewing households (to 2 decimals)? hours c. What is the probability that a household views television more than hours a day (to 4 decimals)?

Answers

Answer 1

Answer:

Answer:

(a) The probability that a household views television between 3 and 9 hours a day is 0.5864.

(b) The viewing hours in the top 2% is 13.49 hours.

Step-by-step explanation:

Let X = daily viewing time of of television hours per household.

The mean daily viewing time is, μ = 8.35 hours.

The standard deviation of daily viewing time is, σ = 2.5 hours.

The random variable X is Normally distributed.

To compute the probability of a Normal random variable, first we need to compute the raw scores (X) to z-scores (Z).

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

(a)

Compute the probability that a household views television between 3 and 9 hours a day as follows:

$P(3<X<9)=P((3-8.35)/(2.5)<(X-\mu)/(\sigma)<(9-8.35)/(2.5))$

$=P(-2.14<Z<0.26)\n=P(Z<0.26)-P(Z<-2.14)\n=0.60257-0.01618\n=0.58639\n\approx0.5864$

Thus, the probability that a household views television between 3 and 9 hours a day is 0.5864.

(b)

Let the viewing hours in the top 2% be denoted by x.

Then,

P (X > x) = 0.02

⇒ P (X < x) = 1 - 0.02

P (X < x) = 0.98

⇒ P (Z < z) = 0.98

The value of z for the above probability is:

z = 2.054

*Use a z-table for the value.

Compute the value of x as follows:

$z=(x-\mu)/(\sigma)\n2.054=(x-8.35)/(2.5)\nx=8.35+(2.054* 2.5)\nx=13.485\nx\approx13.49$

Thus, the viewing hours in the top 2% is 13.49 hours.

(c)

Compute the probability that a household views television more than 5 hours a day as follows:

$P(X>5)=P((X-\mu)/(\sigma)>(5-8.35)/(2.5))$

$=P(Z>-1.34)\n=P(Z<1.34)\n=0.90988\n\approx0.9099$

Thus, the probability that a household views television more than 5 hours a day is 0.9099.

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Answers

Answer:

Step-by-step explanation:

Which situation is most likely to have a constant rate of change?
HELP

Answers

Answer:

the answer i would go with is A

Good luck on your Test :)

Step-by-step explanation:

B doesnt really have a constant rate of change as it depends on how many games happen and usually the longer an arena stays open has no correlation on how many people attend the games there

C has no real constant rate of change as it always ends up stopping after a little bit, and the change is usually not a constant one

D this could count, but since its a number that would go down if its not brought back up, its not a real constant rate of change, since it cant go below or above a certain range

so by process of elimination, A is the answer. also seeing as how its saying the distance with the number of times, that means that its an objective thing, as a track is a set distance, and the distance of a run or the track cant be affected by time or anything and could technically never end. so its a constant thing, meaning the longer the distance is, the higher the laps around the track are, and it could theoretically go on forever.

i hope this helped answer your question! :)

The answer is the question is most definitely A

-4= -(x-8)
negative four equals negative (parentheses x minus 8 parentheses)

Answers

-4=-x-8
+4. +4
—————
0=-x-4
x=-4

Which relationship matches the graph A- {(2,5), (3,3),(1,0),(2,1),(8,8)}

B - {(2,5),(3,3),(0,1),(2,1),(8,8)}

C- {(5,2),(3,3),(0,1),(2,1),(8,8)}

D - {(5,2),(3,3),(1,0),(2,1),(8,8)}