Question 1 a
1 answer:
Answer:
B) A one-sample t-test for population mean would be used.
Step-by-step explanation:
The complete question is shown in the image below.
The marketing executive is interested in comparing the mean number of sales of this year to that of previous year.
The marketing executive already has the value of mean from previous year and uses a sample to calculate the mean and standard deviation of sales for the current year.
Since, data is being collected for one sample only this limits us to chose between one sample test for mean. So now the possible options are one sample t-test for population mean and one sample t-test for population mean.
If we read the statement we can see that we have the value of sample mean and sample standard deviation. Value of population standard deviation is unknown. In cases where value of population standard deviation is not known and sample standard deviation is given, t-test is used.
Therefore, we can conclude that A one-sample t-test for population mean would be used.
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The selection of r objects out of n is done in
many ways.
The total number of selections 10 that we can make from 6+7=13 students is
thus, the sample space of the experiment is 286
A.
<span>"The probability that a randomly chosen team includes all 6 girls in the class."
total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.
</span>
<span>
P(all 6 girls chosen)=35/286=0.12
B.
"</span>The probability that a randomly chosen team has 3 girls and 7 boys.<span>"
with the same logic as in A, the number of groups were all 7 boys are in, is
</span>
<span>
so the probability is 20/286=0.07
C.
"</span>The probability that a randomly chosen team has either 4 or 6 boys.<span>"
case 1: the team has 4 boys and 6 girls
this was already calculated in part A, it is </span>0.12.
<span>
case 2, the team has 6 boys and 4 girls.
there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls.
</span>
<span>
the probability is 105/286=0.367
since case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:
0.12+0.367=0.487 (approximately = 0.49)
D.
"</span><span>The probability that a randomly chosen team has 5 girls and 5 boys.</span><span>"
selecting 5 boys and 5 girls can be done in
</span>
many ways,
so the probability is 126/286=0.44
The total number of selections 10 that we can make from 6+7=13 students is
thus, the sample space of the experiment is 286
A.
<span>"The probability that a randomly chosen team includes all 6 girls in the class."
total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.
</span>
<span>
P(all 6 girls chosen)=35/286=0.12
B.
"</span>The probability that a randomly chosen team has 3 girls and 7 boys.<span>"
with the same logic as in A, the number of groups were all 7 boys are in, is
</span>
<span>
so the probability is 20/286=0.07
C.
"</span>The probability that a randomly chosen team has either 4 or 6 boys.<span>"
case 1: the team has 4 boys and 6 girls
this was already calculated in part A, it is </span>0.12.
<span>
case 2, the team has 6 boys and 4 girls.
there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls.
</span>
<span>
the probability is 105/286=0.367
since case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:
0.12+0.367=0.487 (approximately = 0.49)
D.
"</span><span>The probability that a randomly chosen team has 5 girls and 5 boys.</span><span>"
selecting 5 boys and 5 girls can be done in
</span>
many ways,
so the probability is 126/286=0.44