Answer:
same i feel bad but im here too
Answer: D. This was a random sample. It may have included anyone in attendance.
Step-by-step explanation:
The options are:
A. This was a biased sample. Jim should interview all in attendance.
B. This was a census. Any guest may have participated.
C. This was a random sample. It may not have included anyone in attendance.
D. This was a random sample. It may have included anyone in attendance.
A random sampling is simply referred to as a subset of individuals that are picked from a larger set of individuals.
With regards to the question, Jim wanted to find out what the audience thought about the debate and after the event, he stood at the exit to survey every fifth guest.
This means that it was a random sampling and anyone could have been picked, the sampling wasn't bias.
Let p be
the population proportion. <span>
We have p=0.60, n=200 and we are asked to find
P(^p<0.58). </span>
The thumb of the rule is since n*p = 200*0.60
and n*(1-p)= 200*(1-0.60) = 80 are both at least greater than 5, then n is
considered to be large and hence the sampling distribution of sample
proportion-^p will follow the z standard normal distribution. Hence this
sampling distribution will have the mean of all sample proportions- U^p = p =
0.60 and the standard deviation of all sample proportions- δ^p = √[p*(1-p)/n] =
√[0.60*(1-0.60)/200] = √0.0012.
So, the probability that the sample proportion
is less than 0.58
= P(^p<0.58)
= P{[(^p-U^p)/√[p*(1-p)/n]<[(0.58-0.60)/√0...
= P(z<-0.58)
= P(z<0) - P(-0.58<z<0)
= 0.5 - 0.2190
= 0.281
<span>So, there is 0.281 or 28.1% probability that the
sample proportion is less than 0.58. </span>
