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uranmaximum [27]
3 years ago
5

In a certain county in​ 2013, it was thought that​ 50% of men 50 years old or older had never been screened for prostate cancer.

Suppose a random sample of 200 of these men shows that 120 of them had never been screened.
a. What is the observed frequency of men who said they had not been​ screened?b. What is the observed proportion of men who said they had not been​ screened?c. What is the expected number in the sample to say they had never been screened if​ 50% is the correct​ rate?
Mathematics
1 answer:
joja [24]3 years ago
8 0

Answer:

120,0.60,100

Step-by-step explanation:

Given that in a certain county in​ 2013, it was thought that​ 50% of men 50 years old or older had never been screened for prostate cancer.

Suppose a random sample of 200 of these men shows that 120 of them had never been screened.

a) the observed frequency = 120

b) The observed proportion of men who said they had not been​ screened

=\frac{120}{200} \\=0.60

c) Expected number in the sample to say they had never been screened if​ 50% is the correct​ rate

= np

= 200*0.50=100

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Answer:

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Step-by-step explanation:

Lets mother's  "BROWN" is  "BROWN-M",  

mother's "BLUE"  is  " BLUE-M"

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The kid can have the genotype as follows (list of possible outcomes) :

1. BROWN-M>BROWN-F   ( received BROWN as from mother as from father)

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b) As we can see in a) only 1 outcome from 4 is BLUE-BLUE.  So the probability of BLUE-BLUE genotype is

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So probability that eyes are brown is P(Brown eyes)=3/4 =0.75

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