Answer:
"impossible"
Step-by-step explanation:
I think. :/
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>
Answer:
False
Step-by-step explanation:
It only cuts the risk as stated and other factors such as lifestyle, age, bloodpressure and past medical background also have an impact so you can still have heart issues.
Answer:
Zero, based on the information provided.
Step-by-step explanation:
The output rate of the teller machine is (1 transaction/6 minutes). The input rate is (1 customer/10 minutes). This means that the machine completes a cycle faster than the customers arrive, on the average. I don't know how an average can be calculated without more information. If we assume customers arrive every 10 minutes, and no one screws up the machine, that there should be no waiting line. Is there more information about when the customers arrive? E.g., 50 arrive in the first hour the machine is open.
1/2 = 3/6
Each person would have 1/6 of rice.
