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Darina [25.2K]
3 years ago
12

Suppose babies born in a large hospital have a mean weight of 4022 grams, and a standard deviation of 266 grams. If 53 babies ar

e sampled at random from the hospital, what is the probability that the mean weight of the sample babies would differ from the population mean by more than 45 grams?
Mathematics
1 answer:
irina [24]3 years ago
4 0

Answer: 0.866

Step-by-step explanation:

Mean = 4022

SD = 266

We are looking for the probability of samples 45g differ than the mean. Hence 2 probability combination:

X>4067 and X<3977

Calculate for Z we'll get Z = 0.16917 and -0.161917

Using Z formula we can find P(Z=0.17) = 0.433

Since both probability will occupy the same area, total probability is 0.433x2 = 0.866

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1) Proportion a.

7 : 28 = 2 : 8

That means that the ratio 7 / 28 is equal to the ratio 2 : 8. 

You can verify the equality by simplifying to the simplest form:

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In words that is 7 is to 28 as 1 is to 4 ← answer


2) Proportion b.<span>. 3⁄1 = 18⁄6
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</span><span>In words: 3 is to q as 18 is to 6 ← answer
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</span><span>You can verify the proportion by simplifiying the second fraction:
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</span><span>     18 / 6 = [18 / 6] / [ 6 / 6] = 3 / 1
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</span><span>3) Proportion c. 9 : 72 = 2 : 16
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</span><span>In words: 9 is to 72 as 2 is to 16 ← answer
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</span><span>Proove the proportion:
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</span><span>4) Proportion d. 81⁄9 = 45⁄5</span>

In words: 81 is to 9 as 45 is to 5 ← answer


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