Marley drives to work every day and passes two independently operated traffic lights. The probability that both lights are red i
2 answers:
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Answer:
Step-by-step explanation:
For the random variable we define the possible values for this variable on this case
. We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:
We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable like this:
For the distribution of we need to take in count that we are finding a conditional distribution.
given
, for this case we see that
, so then exist
ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined:
And if we want to find the joint probability we just need to do this:
And if we multiply the probabilities founded we got:
Answer:
(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].
(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval
(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.
Step-by-step explanation:
We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.
Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;
P.Q. = ~ N(0,1)
where, = sample proportion of Americans who decide to not go to college = 48%
n = sample of American adults = 331
p = population proportion of Americans who decide to not go to
college because they cannot afford it
<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>
<em />
<u>So, 90% confidence interval for the population proportion, p is ;</u>
P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level
of significance are -1.645 & 1.645}
P(-1.645 < < 1.645) = 0.90
P( <
<
) = 0.90
P( < p <
) = 0.90
<u>90% confidence interval for p</u> = [ ,
]
= [ ,
]
= [0.4348, 0.5252]
(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].
(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.
3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.
So, the margin of error =
= 54.79
n =
n = 3001.88 ≈ 3002
Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.
Answer:
£265
Step-by-step explanation:
Total fraction of money spent=
=
Actual money spent = * £636
= £371
Money he has left = £(636 - 371)
= £265