I need help ASAP please
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Answer:
(a) P (p' > 0.25) = 0.
(b) P (0.15 < p' < 0.20) = 0.7784.
(c) P (133 < X < 171) = 0.2205
Step-by-step explanation:
Let <em>p</em> = proportion of business trip that were for internal company visit.
It is provided that 19% responded that it was for an internal company visit, i.e.
<em>p</em> = 0.19.
A sample of <em>n</em> = 950 business travelers are randomly selected.
According to the Central limit theorem, if from an unknown population large samples of sizes <em>n</em> > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of this sampling distribution of sample proportion is:
The standard deviation of this sampling distribution of sample proportion is:
Thus, the distribution of is N (<em>μ</em> = 0.19, <em>σ </em>= 0.013).
(a)
Compute the probability that more than 25% of the business travelers say that the reason for their most recent business trip was an internal company visit as follows:
P (p' > 0.25) = P (Z > 4.62) = 0
*Use a <em>z</em>-table for the probability.
Thus, the probability that more than 25% of the business travelers say that the reason for their most recent business trip was an internal company visit is 0.
(b)
Compute the probability that between 15% and 20% of the business travelers say that the reason for their most recent business trip was an internal company visit as follows:
P (0.15 < p' < 0.20) = P (-3.08 < Z < 0.77)
= P (Z < 0.77) - P (Z < -3.08)
= 0.7794 - 0.0010
= 0.7784
*Use a <em>z</em>-table for the probability.
Thus, the probability that between 15% and 20% of the business travelers say that the reason for their most recent business trip was an internal company visit is 0.7784.
(c)
Compute the proportion of <em>X</em> = 133 and <em>X</em> = 171 as follows:
p' = 133/950 = 0.14
p' = 171/950 = 0.18
Compute the value of P (0.14 < p' < 0.20) as follows:
P (0.14 < p' < 0.18) = P (-3.85< Z < -0.77)
= P (Z < -0.77) - P (Z < -3.85)
= 0.2206- 0.0001
= 0.2205
*Use a <em>z</em>-table for the probability.
Thus, the probability that between 133 and 171 of the business travelers say that the reason for their most recent business trip was an internal company visit is 0.2205.
So, she has 3hrs to grade all papers, for 35 students.. alrite.
the first 5, she does them in 30minutes.. what's the speed rate? well, 5/30 or 1/6
now, she has still 2 hours and a half, or 150 minutes, to do the remaining 30 papers... she has to work at a rate of 30/150 then... which is 1/5 simplified.
now if we take 1/6 as the 100%, what is 1/5 in percentage then?
so 1/5 is 120% in relation to 1/6... meaning the rate of 1/5 she needs to move through, namely 1 paper every 5 minutes, is 20% faster than 1/6.
the first 5, she does them in 30minutes.. what's the speed rate? well, 5/30 or 1/6
now, she has still 2 hours and a half, or 150 minutes, to do the remaining 30 papers... she has to work at a rate of 30/150 then... which is 1/5 simplified.
now if we take 1/6 as the 100%, what is 1/5 in percentage then?
so 1/5 is 120% in relation to 1/6... meaning the rate of 1/5 she needs to move through, namely 1 paper every 5 minutes, is 20% faster than 1/6.