Ashley opened a lemonade stand on her street. It costs her $12.50 to buy the supplies each day, and she sells lemonade for $2.50
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Answer:
The p-value of the test is 0.1469 > 0.05, which means that there is no reason to believe that the proportion of adults favoring capital punishment today has increased, using a 0.05 level of significance.
Step-by-step explanation:
Suppose that, in the past, 40% of all adults favored capital punishment. Test if the proportion has increased:
At the null hypothesis, we test if the proportion is still of 40%, that is:
At the alternative hypothesis, we test if the proportion has increased, that is, is greater than 40%, so:
The test statistic is:
In which X is the sample mean, is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
0.4 is tested at the null hypothesis:
This means that
Random sample of 15 adults, 8 favor capital punishment.
This means that
Value of the test statistic:
P-value of the test and decision:
The p-value of the test is the probability of finding a sample proportion of 0.5333 or more, which is 1 subtracted by the p-value of z = 1.05.
Looking at the z-table, z = 1.05 has a p-value of 0.8531.
1 - 0.8531 = 0.1469.
The p-value of the test is 0.1469 > 0.05, which means that there is no reason to believe that the proportion of adults favoring capital punishment today has increased, using a 0.05 level of significance.
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%