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qwelly [4]
2 years ago
9

A candidate for mayor wants to know how strong her support is among voters, so she posts an online survey on her campaign websit

e for voters to take. 
Is this sample of voters likely to be biased or unbiased?   A. biased   B. unbiased   To determine whether voters would be willing to vote for a tax increase to improve the city's schools, the board of education surveys 500 parents of children in the city's schools. 
Why is this sample likely to be biased?   A. The entire population of the city was not surveyed.   B. Parents of school children are too busy to be bothered with surveys.   C. There are not enough people in the sample.   D. Parents of school children are more likely to be willing to pay more to improve schools.  In a bin containing 500 marbles, 180 are red, 240 are blue, and 80 are green. In a bowl of 25 marbles scooped from this bin, how many would be expected to be green?   A. 4   B. 8   C. 9   D. 12   A battery manufacturer randomly tests 500 batteries and finds that 3 are defective. 
How many defective batteries would be expected in a shipment of 12,000 batteries?   A. 120   B. 72   C. 60   D. 36                                                                                 A large bin of mixed nuts contains peanuts, cashews, almonds, and walnuts. Sheila scoops out a bowl full of nuts and counts how many of each type she has: 17 peanuts, 9 cashews, 12 almonds, and 10 walnuts. 
Armando scoops out a bowl of 32 nuts. About how many should he expect to be cashews?   A. 5   B. 6   C. 9   D. 18
Mathematics
1 answer:
maks197457 [2]2 years ago
3 0
1) A. Biased because the population could be uneven. Too many of one ethnicity, too many of one gender, etc. 

2) I believe this one is C. because there are not enough people. You need to survey the majority of parents instead of a selected few. 

3) There are 80 green marbles per 500, therefore calculate the percentage. \frac{80}{500}=0.16 or 16 percent. Multiply 16 percent or 0.16 times 25. It gives you 4.

4) This one is a probability thing too. For every 500 batteries, there are 3 that are defective, therefore, \frac{3}{500}= 0.006. 

0.006*12000 = 72

5) Yet another one. Add up all the nuts and you get 48. The chance of getting a cashew from these 48 is: \frac{9}{48}=0.1875. 

Therefore: 0.1875*32=6
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A scientist claims that 7% 7 % of viruses are airborne. If the scientist is accurate, what is the probability that the proportio
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Answer:

The probability is P(|p-\^{p}| >  0.03)  =   0.0040

Step-by-step explanation:

From the question we are told that

   The population proportion is p =  0.07

   The mean of the sampling distribution is   \mu_p =  0.07

   The sample size is n = 600

Generally the standard deviation is mathematically represented as

     \sigma_p  =  \sqrt{\frac{p (1 -p)}{n} }

=>    \sigma_p  =  \sqrt{\frac{0.07(1 -0.07)}{600} }    

=>    \sigma_p  =  0.010416    

Generally the probability that the proportion of airborne viruses in a sample of 600 viruses would differ from the population proportion by greater than 3% is mathematically represented as

      P(|p-\^{p}| >  0.03) =  1 - P(|p -\^{p}| \le 0.03)

=>   P(|p-\^{p}| >  0.03)  =  1 -  P(-0.03 \le p -\^{p} \le 0.03 )

Now  add p  to  both side of the inequality

=>   P(|p-\^{p}| >  0.03)  =  1 -  P( 0.07-0.03  \le \^{p} \le 0.03+ 0.07 )

=>   P(|p-\^{p}| >  0.03)  =  1 -  P(0.04 \le \^{p} \le 0.10 )

Now  converting the probabilities to their respective standardized score

=> P(|p-\^{p}| >  0.03)  =  1 -  P(\frac{0.04 - 0.07}{0.010416}  \le Z \le \frac{0.10 -0.07}{0.010416}  )

=> P(|p-\^{p}| >  0.03)  =  1 -  P(-2.88  \le Z \le 2.88 )

=>   P(|p-\^{p}| >  0.03)  =   1 - [P(Z \le 2.88) - P(Z \le -2.88)]

From the z-table  

       P(Z \le 2.88)  =  0.9980

and

       P(Z \le -2.88)  = 0.0020

So

     P(|p-\^{p}| >  0.03)  =   1 - [0.9980 - 0.0020]

=>   P(|p-\^{p}| >  0.03)  =   0.0040

     

7 0
3 years ago
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