DONT ANSWER JUST TELL ME HOW TO DO IT!!!!! 15 POINTS!!! since there is two numbers infant of the parentheses which one do you di
stribute???
DONT REPORT PEOPLE WHO GIVE HINTS INSTEAD OF ANSWERS IT IS RUDE!!!
DONT REPORT PEOPLE WHO GIVE HINTS INSTEAD OF ANSWERS IT IS RUDE!!!
2 answers:
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Answer:
a) 59.34%
b) 44.82%
c) 26.37%
d) 4.19%
Step-by-step explanation:
(a)
There are in total <em>4+5+6 = 15 bulbs</em>. If we want to select 3 randomly there are K ways of doing this, where K is the<em> combination of 15 elements taken 3 at a time </em>
As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.
So, the probability of selecting exactly 2 bulbs of 75 W is
(b)
The probability of selecting three 40-W bulbs is
The probability of selecting three 60-W bulbs is
The probability of selecting three 75-W bulbs is
Since <em>the events are disjoint</em>, the probability of taking 3 bulbs of the same kind is the sum 0.0527+0.1318+0.2637 = 0.4482 = 44.82%
(c)
There are 6*5*4 ways of selecting one bulb of each type, so the probability of selecting 3 bulbs of each type is
(d)
The probability that it is necessary to examine at least six bulbs until a 75-W bulb is found, <em>supposing there is no replacement</em>, is the same as the probability of taking 5 bulbs one after another without replacement and none of them is 75-W.
As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is
Since there are no replacement, the probability of taking a second non 75-W bulb is now
Following this procedure 5 times, we find the probabilities
which are
0.6, 0.5714, 0.5384, 0.5, 0.4545
As the events are independent, the probability of choosing 5 non 75-W bulbs is the product
0.6*0.5714*0.5384*0.5*0.4545 = 0.0419 = 4.19%
Answer:
(E) The bias will decrease and the variance will decrease.
Step-by-step explanation:
Given that researchers working the mean weight of a random sample of 800 carry-on bags to e the airline.
We have to find out the effect of increasing the sample size on variance and bias.
We know as per central limit theorem, sample mean follows a normal distribution with mean = sample mean
and std deviation of sample mean = std error =
Thus std error the square root of variance is inversely proportional to the square root of sample size.
Also whenever we increase sample size the chances of bias would decrease as the sample size becomes larger
So correct answer is both bias and variation will decrease.
(E) The bias will decrease and the variance will decrease.