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ratelena [41]
3 years ago
13

Suppose that during the 1990s, the population of a certain country was increasing by 1.7% each year. If the population at the en

d of 1993 was 5.4 million, what was the population at the end of 1994?
a.
91,800 people
c.
5,400,000 people
b.
5,612,879 people
d.
5,491,800 people
Mathematics
1 answer:
Sidana [21]3 years ago
6 0

Answer:

If the population increases by 1.7% every year then

1994 population = 1993 population * .017 + 1993 population

1994 population = 5,491,800

Answer is "d".

Step-by-step explanation:

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Step-by-step explanation:

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Newborn babies: A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights
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Answer:

a) 615

b) 715

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Step-by-step explanation:

According to the Question,

  • Given that,  A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 732 babies born in New York. The mean weight was 3311 grams with a standard deviation of 860 grams

  • Since the distribution is approximately bell-shaped, we can use the normal distribution and calculate the Z scores for each scenario.

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Now,

For x = 4171,  Z = (4171 - 3311)/860 = 1  

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Next, multiply that by the sample size of 732.

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  • For part b, use the same method except x is now 1591.    

Z = (1581 - 3311)/860 = -2    

  • P(Z > -2) , using the Z table is 1 - 0.0228 = 0.9772 . Now 732(0.9772) = 715.3104, so approximately 715 will weigh more than 1591.

 

  • For part c, we now need to get two Z scores, one for 3311 and another for 5031.

Z1 = (3311 - 3311)/860 = 0

Z2 = (5031 - 3311)/860= 2  

P(0 ≤ Z ≤ 2) = 0.9772 - 0.5000 = 0.4772

  approximately 47% fall between 0 and 1 standard deviation, so take 0.47 times 732 ⇒ 732×0.47 = 344.

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