Answer:
The mean and the standard deviation of the sampling distribution of the number of students who preferred to get out early are 0.533 and 0.82
Step-by-step explanation:
According to the given data we have the following:
Total sample of students= 150
80 students preferred to get out 10 minutes early
Therefore, the mean of the sampling distribution of the number of students who preferred to get out early is = 80/150 = 0.533
Therefore, standard deviation of the sampling distribution of the number of students who preferred to get out early= phat - p0/sqrt(p0(1-p)/)
= 0.533-0.5/sqrt(0.5*0.5/15))
= 0.816 = 0.82
Answer:
20.5 is the answer!! all you have to do is add em.
Answer:
The probability that at least 280 of these students are smokers is 0.9664.
Step-by-step explanation:
Let the random variable <em>X</em> be defined as the number of students at a particular college who are smokers
The random variable <em>X</em> follows a Binomial distribution with parameters n = 500 and p = 0.60.
But the sample selected is too large and the probability of success is close to 0.50.
So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:
1. np ≥ 10
2. n(1 - p) ≥ 10
Check the conditions as follows:

Thus, a Normal approximation to binomial can be applied.
So,

Compute the probability that at least 280 of these students are smokers as follows:
Apply continuity correction:
P (X ≥ 280) = P (X > 280 + 0.50)
= P (X > 280.50)

*Use a <em>z</em>-table for the probability.
Thus, the probability that at least 280 of these students are smokers is 0.9664.
Answer:
This is matter of ratio!
42/150 = x/120. It's 33.6 grams :D
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Multiplicity of a root means, is there that many times.