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ArbitrLikvidat [17]
3 years ago
14

A professor wants to determine information about the Internet usage of middle school students in one state. A sample of 500 boys

and girls from middle schools across the state were asked, “How do you access the Internet when you are not at school?” Which best explains why the sample can be used to determine statistics that apply to the population?
A. The sample represents the population
B. The sample is larger than the population
C. Each person in the population knows at least one person included in the example
D. Each person in the sample has internet access while they are not at school

Please answer ASAP!
Mathematics
2 answers:
ICE Princess25 [194]3 years ago
6 0

Answer:

The answer is A

Step-by-step explanation:


zlopas [31]3 years ago
4 0
The answer is A because a sample of 500 boys and girls from middle schools across the state represents the population.
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An electronics hobbyist has three electronic parts cabinets with two drawers each.
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Answer:

a) 0.5 = 50% probability that an NPN transistor will be selected.

b) 0.3333 = 33.33% probability that it came from the cabinet that contains both types

c) 66.67% probability that it comes from the cabinet that contains only NPN transistors

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

a) What is the probability that an NPN transistor will be selected?

1/3 probability that the first cabinet is chosen. This cabinet has two transistors, both of which are NPN, so 100% probability of selecting a NPN transistor.

1/3 probability that the second cabinet is chosen. This cabinet has two transistors, both of which are PNP, so 0% probability of selecting a NPN transistor.

1/3 probability that the second cabinet is chosen. This cabinet has two transistors, one of which is NPN, so 50% probability of selecting a NPN transistor.

So

p = \frac{1}{3}*1 + \frac{1}{3}*0 + \frac{1}{3}*0.5 = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = 0.5

0.5 = 50% probability that an NPN transistor will be selected.

b) Given that the hobbyist selects an NPN transistor, what is the probability that it came from the cabinet that contains both types?

Here we use the conditional probability formula.

Event A: NPN transistor

Event B: From the third cabinet.

50% probability that an NPN transistor will be selected, so P(A) = 0.5.

1/6 probability that it is from the third cabinet and NPN, so P(A \cap B) = \frac{1}{6}

The desired probability is:

P(B|A) = \frac{\frac{1}{6}}{0.5} = 0.3333

0.3333 = 33.33% probability that it came from the cabinet that contains both types.

c) Given that an NPN transistor is selected what is the probability that it comes from the cabinet that contains only NPN transistors?

Either it comes from the cabinet with only NPN transistors, or it comes from the cabinet with both types of transistors. The sum of the probabilities of these outcomes is 100%. So

x + 33.33 = 100

x = 66.67

66.67% probability that it comes from the cabinet that contains only NPN transistors

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