Answer:
A recent Harris poll asked a random sample of 1016 adults nation-wide whether or not they smoked cigarettes. 22% said they smoked. Based on this sample, can you conclude that...
a.) The expected value for the percent of all adults world-wide who would say they smoked cigarettes is 22%?
b). The expected value for the percent of students at UI who would say they smoked cigarettes is 22%?
c). The expected value for the percent of professors at UI who would say they smoked cigarettes is 22%?
d). The expected value for the percent of all US adults who would say they smoked cigarettes is 22%?
The correct answer is
d).The expected value for the percent of all US adults who would say they smoked cigarettes is 22%?
Step-by-step explanation:
The expected value is arrived at by finding the product of a possible output and the probability that the output will occur and summing up the results. Expected value can be used for investment management to calculate options and make decisions most likely to bring about the desired gain. The random variable provides categorisation of the outcomes of the game while the expected provides the probability of aan outcome
In the above, the source of the sample is nationwide whereby 22 % said they smoked therefore it cannot be applied to a different population that has a different expected value for the same survey.
Answer:
x= -3+√2 or x= -3-√2 ( answer : A and B )
Step-by-step explanation:
hello :
x²+6x+9 = (x+3)².....identity
x²+6x+9 =2 means : (x+3)²=2
so : (x+3 = √2) or (x+3 = - √2)
so two solutions :
x= -3+√2 or x= -3-√2
Answer:
the answer to the question is y=5*
Answer:
0.1994 is the required probability.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 166 pounds
Standard Deviation, σ = 5.3 pounds
Sample size, n = 20
We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.
Formula:
Standard error due to sampling =
P(sample of 20 boxers is more than 167 pounds)
Calculation the value from standard normal z table, we have,
0.1994 is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds
