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s344n2d4d5 [400]
3 years ago
14

The article "High Cumulative Risk of Lung Cancer Death among Smokers and Nonsmokers" (P. Brennan, et al. American Journal of Epi

demiology , 2006:1233–1241) states that the probability is 0.24 that a man who is a heavy smoker will contract lung cancer. True or false:
a. In a sample of 100 men who are heavy smokers, exactly 24 of them will contract lung cancer.
b. In a sample of 100 men who are heavy smokers, the number who will contract lung cancer is likely to be close to 24, but not exactly equal to 24.
c. As more and more heavy-smoking men are sampled, the proportion who contract lung cancer will approach 0.24.
Mathematics
1 answer:
Dmitry [639]3 years ago
7 0

Answer:

a)False: In a sample of 100 men who are heavy smokers, exactly 24 of them will contract lung cancer.

b)True: In a sample of 100 men who are heavy smokers, the number who will contract lung cancer is likely to be close to 24.

c)True: As more and more heavy-smoking men are sampled, the proportion who contract lung cancer will approach 0.24. Circle one: TRUE FALSE

You might be interested in
Are 5(2x-y) and 15x-5y equivalent expressions?
krek1111 [17]

No, they are not equivalent.

5(2x - y) = 10x - 5y and it is not equal to 15x - 5y.

(using distribution property)

6 0
3 years ago
The following probability distributions of job satisfaction scores for a sample of information systems (IS) senior executives an
Mashcka [7]

Answer:

Step-by-step explanation:

To calculate ;

1) the expected value of the job satisfaction score for senior executives ;

expected value = Summation (Px)

= 1 x 0.05 + 2 x 0.09 + 3 x 0.03 + 4 x 0.42 + 5 x 0.41

= 4.05

2) the expected value of the job satisfaction score for middle managers;

= 1 x 0.04 + 2 x 0.10 + 3 x 0.12 + 4 x 0.46 + 5 x 0.28

= 3.84

c) the variance of job satisfaction scores for executives and middle managers (to 2 decimals).

Executives ; Variance = Summation(PX^2 - Summation(PX)^2

i) For Executive Managers = 1 x 0.05 + 2^2 x 0.09 + 3^2 x 0.03 + 4^2 x 0.42 + 5^2 x 0.41 - 4.05^2 = 1.246 = 1.25

ii) for middle managers ; 1 x 0.04 + 2^2 x 0.10 + 3^2 x 0.12 + 4^2 x 0.46 + 5^2 x 0.28 - 3.84^2 = 1.134 = 1.13

d) the standard deviation of job satisfaction scores for both probability distributions (to 2 decimals). Executives, Middle managers;

For Executives = square root [ 1 x 0.05 + 2^2 x 0.09 + 3^2 x 0.03 + 4^2 x 0.42 + 5^2 x 0.41 - 4.05^2] = 1.12

For Middle Managers ; Square root [1 x 0.04 + 2^2 x 0.10 + 3^2 x 0.12 + 4^2 x 0.46 + 5^2 x 0.28 - 3.84^2 ] = 1.06

e) from the values gotten for the variance of both executive and middle managers, the variance of the former is more than that of the latter as such higher satisfaction with the executive managers.

5 0
3 years ago
The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability that
tekilochka [14]

Answer:

a) P(X∩Y) = 0.2

b) P_1 = 0.16

c) P = 0.47

Step-by-step explanation:

Let's call X the event that the motorist must stop at the first signal and Y the event that the motorist must stop at the second signal.

So, P(X) = 0.36, P(Y) = 0.51 and P(X∪Y) = 0.67

Then, the probability P(X∩Y) that the motorist must stop at both signal can be calculated as:

P(X∩Y) = P(X) + P(Y) - P(X∪Y)

P(X∩Y) = 0.36 + 0.51 - 0.67

P(X∩Y) = 0.2

On the other hand, the probability P_1 that he must stop at the first signal but not at the second one can be calculated as:

P_1 = P(X) - P(X∩Y)

P_1 = 0.36 - 0.2 = 0.16

At the same way, the probability P_2 that he must stop at the second signal but not at the first one can be calculated as:

P_2 = P(Y) - P(X∩Y)

P_2 = 0.51 - 0.2 = 0.31

So, the probability that he must stop at exactly one signal is:

P = P_1+P_2\\P=0.16+0.31\\P=0.47

7 0
3 years ago
PLEASE HELP ME!!
Dmitry_Shevchenko [17]
I think it’s A sorry if it’s wrong!
6 0
3 years ago
A 1.5m wire carries a 7 A current when a potential difference of 54 V is applied. What is the resistance of the wire?
borishaifa [10]

Answer:

R = 7.71 ohms

Step-by-step explanation:

By ohm's law: R= V/I

where,  R = resistance of wire, V = potential difference, I = electric current

Given: V = 54 and I = 7 A

Plug in the above values in the ohm's law, we get

R = 54 V/7 A

R = 7.71 ohms

Hope this will helpful.

Thank you.

6 0
3 years ago
Read 2 more answers
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