Three national news networks are trying to estimate what percentage of
american voters support a particular presidential candidate. to estimate the
population parameter, each of the three networks collect data from a simple
random sample of voters. here are the numbers of voters in the networks
random samples:
• network a: 1500 voters sampled
• network b: 1250 voters sampled
• network c: 1800 voters sampled
which sample is likely to have the least variability?
o
a. the network a sample
o
b. the network b sample
o
c. the network c sample

Explanation:

The variability of a sample may be described by using some common statistics like the variance and the standard deviation.

The variance and the standard deviation measure the difference between the values of the data  and the mean. The variance is the average square of those differences or deviations. The standard deviation is the square root of the variance.

Thus, both variance and deviations, being an average of the differences of the data from the mean contain the number of data in the denominator of the equation.

That lets you know to conclude that the larger the number of data (the larger the sample) the smaller the variance and the standard deviation, this is the lower the variability of the data.

Hence, Network C, which has the greatest number of voters sampled (1,800) is likely to have the least variability.

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %

idk

step-by-step explanation:

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