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Serhud [2]
2 years ago
13

A random survey of seventh and eighth grade students was conducted to find out how many hours are spent doing homework each nigh

t. Of the two sets of data, which has the greater median?
A: seventh graders
B: eighth graders
C: they have the same median
D: impossible to determine from the box plot

Mathematics
1 answer:
sammy [17]2 years ago
4 0
Eight graders is the answer
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Customers are used to evaluate a preliminary product design. In the past, 95% of highly successful products received good review
Sever21 [200]

Answer:

a. 61.5%; b. About 61.8%; c. About 36.4%

Step-by-step explanation:

This is a kind of question that we can solve using the Bayes' Theorem. We have here all the different conditional probabilities we need to solve this problem.

According to that theorem, the probability of a selected product attains a good review is:

\\ P(G) = P(G|H)*P(H) + P(G|M)*P(M) + P(G|P)*P(P) (1)

In words, the probability that a selected product attains a <em>good review</em> is an <em>event </em>that depends upon the sum of the conditional probabilities that the product comes from <em>high successful product</em> P(G|H) by the probability that this product is a <em>highly successful product</em> P(H), plus the same about the rest of the probabilities, that is, P(G|M)*P(M) or the probability that the product has a good review coming from a <em>moderately successful</em> product by the probability of being moderately successful, and a good review coming from a poor successful product by the probability of being poor successful or P(G|P)*P(P).

<h3>The probability that a randomly selected product attains a good review</h3>

In this way, the probability that a randomly selected product attains a good review is the result of the formula (1). Where (from the question):

P(G|H) = 95% or 0.95 (probability of receiving a good review being a highly successful product)

P(G|M) = 60% or 0.60 (probability of receiving a good review being a moderately successful product)

P(G|P) = 10% or 0.10 (probability of receiving a good review being a poorly successful product)

P(H) = 40% or 0.40 (probability of  being a highly successful product).

P(M) = 35% or 0.35 (probability of  being a moderately successful product).

P(P) = 25% or 0.25 (probability of  being a poor successful product).

Then,

\\ P(G) = P(G|H)*P(H) + P(G|M)*P(M) + P(G|P)*P(P)

\\ P(G) = 0.95*0.40 + 0.60*0.35 + 0.10*0.25

\\ P(G) = 0.615\;or\; 61.5\%

That is, <em>the probability that a randomly selected product attains a good review</em> is 61.5%.

<h3>The probability that a new product attains a good review is a highly successful product</h3>

We are looking here for P(H|G). We can express this probability mathematically as follows (another conditional probability):

\\ P(H|G) = \frac{P(G|H)*P(H)}{P(G)}

We can notice that the probability represents a fraction from the probability P(G) already calculated. Then,

\\ P(H|G) = \frac{0.95*0.40}{0.615}

\\ P(H|G) =\frac{0.38}{0.615}

\\ P(H|G) =0.618

Then, the probability of a product that attains a good review is indeed a highly successful product is about 0.618 or 61.8%.

<h3>The probability that a product that <em>does not attain </em>a good review is a moderately successful product</h3>

The probability that a product does not attain a good review is given by a similar formula than (1). However, this probability is the complement of P(G). Mathematically:

\\ P(NG) = P(NG|H)*P(H) + P(NG|M)*P(M) + P(NG|P)*P(P)

P(NG|H) = 1 - P(G|H) = 1 - 0.95 = 0.05

P(NG|M) = 1 - P(G|M) = 1 - 0.60 = 0.40

P(NG|P) = 1 - P(G|M) = 1 - 0.10 = 0.90

So

\\ P(NG) = 0.05*0.40 + 0.40*0.35 + 0.90*0.25

\\ P(NG) = 0.385\;or\; 38.5\%

Which is equal to

P(NG) = 1 - P(G) = 1 - 0.615 = 0.385

Well, having all this information at hand:

\\ P(M|NG) = \frac{P(NG|M)*P(M)}{P(NG)}

\\ P(M|NG) = \frac{0.40*0.35}{0.385}

\\ P(M|NG) = \frac{0.14}{0.385}

\\ P(M|NG) = 0.363636... \approx 0.364

Then, the <em>probability that a new product does not attain a good review and it is a moderately successful product is about </em>0.364 or 36.4%.

8 0
2 years ago
Solve 9/11 + 3/13 × 26/33 - (21/121÷ 71/111)
Julli [10]

Answer:

0.72 or 0.72866953788  0.7  0.72866953788

3 0
2 years ago
When a breeding group of animals is introduced into a restricted area such as a wildlife reserve, the population can be expected
jasenka [17]

Answer:

A. Initially, there were 12 deer.

B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. After 15 years, there will be 410 deer.

D. The deer population incresed by 30 specimens.

Step-by-step explanation:

N=\frac{12.36}{0.03+0.55^t}

The amount of deer that were initally in the reserve corresponds to the value of N when t=0

N=\frac{12.36}{0.33+0.55^0}

N=\frac{12.36}{0.03+1} =\frac{12.36}{1.03} = 12

A. Initially, there were 12 deer.

B. N(10)=\frac{12.36}{0.03 + 0.55^t} =\frac{12.36}{0.03 + 0.0025}=\frac{12.36}{y}=380

B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. N(15)=\frac{12.36}{0.03+0.55^15}=\frac{12.36}{0.03 + 0.00013}=\frac{12.36}{0.03013}= 410

C. After 15 years, there will be 410 deer.

D. The variation on the amount of deer from the 10th year to the 15th year is given by the next expression:

ΔN=N(15)-N(10)

ΔN=410 deer - 380 deer

ΔN= 30 deer.

D. The deer population incresed by 30 specimens.

8 0
2 years ago
Margaret and Bethany have a lemonade stand for one week each summer.
Monica [59]

Answer:

6.00

Step-by-step explanation:

5 0
2 years ago
A community group sells 2,000 tickets for its raffle. The grand prize is a car. Neil and 9 of his friends buy 10 tickets each. W
NeTakaya

Answer:

1/100

Step-by-step explanation:

It's because he and his 9 friends so 10 people each by 10 tickets. Ten times ten is 100 so it is one out of one-hundred.

3 0
2 years ago
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