How to find the expectation and the standard deviation about this question? The 2019 Stack...
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How to find the expectation and the standard deviation about this question?
The 2019 Stack Overflow Developer Survey Results Are InOptimally combining samples to estimate averagesWhat is the probability that at least 4 customers arrived between 9am and 10am?What is the variance of the time of the arrival of the 3rd customer?Confidence interval in comparing service timesConfidence stats questionScaling Random variables by a constantShouldn't sample standard deviation DECREASE with increased sample size?Probability of having at least one table with $n$ customersProbability and expected gainTime evolution of the probability density
$begingroup$
According to the observation, 80% customers order breakfast A, while other customers order breakfast B. At this moment, all six tables in the cafe are occupied, each table with one customer.
Then the question is :breakfast A is cost $45 while breakfast B is cost 38 dollar. What is the expectation and the standard deviation of the revenue for repeated samples of six individual customers?
probability
asked Mar 21 at 4:52
HakzihippoHakzihippo
1
$endgroup$
add a comment |
$begingroup$
According to the observation, 80% customers order breakfast A, while other customers order breakfast B. At this moment, all six tables in the cafe are occupied, each table with one customer.
Then the question is :breakfast A is cost $45 while breakfast B is cost 38 dollar. What is the expectation and the standard deviation of the revenue for repeated samples of six individual customers?
probability
asked Mar 21 at 4:52
HakzihippoHakzihippo
1
$endgroup$
add a comment |
$begingroup$
According to the observation, 80% customers order breakfast A, while other customers order breakfast B. At this moment, all six tables in the cafe are occupied, each table with one customer.
Then the question is :breakfast A is cost $45 while breakfast B is cost 38 dollar. What is the expectation and the standard deviation of the revenue for repeated samples of six individual customers?
probability
asked Mar 21 at 4:52
HakzihippoHakzihippo
1
$endgroup$
According to the observation, 80% customers order breakfast A, while other customers order breakfast B. At this moment, all six tables in the cafe are occupied, each table with one customer.
Then the question is :breakfast A is cost $45 while breakfast B is cost 38 dollar. What is the expectation and the standard deviation of the revenue for repeated samples of six individual customers?
probability
probability
asked Mar 21 at 4:52
HakzihippoHakzihippo
1
asked Mar 21 at 4:52
HakzihippoHakzihippo
1
asked Mar 21 at 4:52
HakzihippoHakzihippo
1
asked Mar 21 at 4:52
HakzihippoHakzihippo
1
asked Mar 21 at 4:52
HakzihippoHakzihippo
1
1
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Every customer can order either A or B with probability 0.8 or 0.2 respectively (lets call these $p_A$ and $p_B = 1 - p_A$). The cost of A is $45 and B is $38 (call these $r_A$ and $r_B$).
Expected value of revenue from one customer (say $r/c$) is,
$$E[r/c] = (p_A times r_A) + (p_B times r_B)$$
$$E[r/c] = (p_A times r_A) + ((1-p_A) times r_B)$$
Expected revenue from 6 customers is just 6 times the expression above. This is also the mean revenue from 6 customers (let's call this $mu$).
As for the standard deviation, it is,
begin{equation}
sigma = sqrt{Var(r)}
end{equation}
where,
$$Var(r) = left(sum Pr(r = r_i) times r_i^2 right) - mu^2$$
Now for a single customer $r_i$ could only be either $r_A$ or $r_B$, but for 6 customers, there are $2^6$ different permutations of meals that can be generated for a given set of customers.
However, if $n_A$ customers chose A and $n_B = 6 - n_A$ customers chose B, then the revenue generated is $n_A times r_A + (6-n_A) times r_B$. As $n_A$ can be anywhere between 0 (no customer chose A) to 6 (all of them chose A). The probability that $n_A$ will choose A and $6-n_A$ choose B is $p_A^{n_A} (1-p_A)^{6-n_A}$.
$$Var(r) = left(sum_{n_A = 0}^{6} p_A^{n_A} (1-p_A)^{6-n_A} (n_A times r_A + (6-n_A) times r_B)^2 right) - mu^2$$
Use this to get the standard deviation.
answered Mar 21 at 8:12
Balakrishnan RajanBalakrishnan Rajan
1519
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Every customer can order either A or B with probability 0.8 or 0.2 respectively (lets call these $p_A$ and $p_B = 1 - p_A$). The cost of A is $45 and B is $38 (call these $r_A$ and $r_B$).
Expected value of revenue from one customer (say $r/c$) is,
$$E[r/c] = (p_A times r_A) + (p_B times r_B)$$
$$E[r/c] = (p_A times r_A) + ((1-p_A) times r_B)$$
Expected revenue from 6 customers is just 6 times the expression above. This is also the mean revenue from 6 customers (let's call this $mu$).
As for the standard deviation, it is,
begin{equation}
sigma = sqrt{Var(r)}
end{equation}
where,
$$Var(r) = left(sum Pr(r = r_i) times r_i^2 right) - mu^2$$
Now for a single customer $r_i$ could only be either $r_A$ or $r_B$, but for 6 customers, there are $2^6$ different permutations of meals that can be generated for a given set of customers.
However, if $n_A$ customers chose A and $n_B = 6 - n_A$ customers chose B, then the revenue generated is $n_A times r_A + (6-n_A) times r_B$. As $n_A$ can be anywhere between 0 (no customer chose A) to 6 (all of them chose A). The probability that $n_A$ will choose A and $6-n_A$ choose B is $p_A^{n_A} (1-p_A)^{6-n_A}$.
$$Var(r) = left(sum_{n_A = 0}^{6} p_A^{n_A} (1-p_A)^{6-n_A} (n_A times r_A + (6-n_A) times r_B)^2 right) - mu^2$$
Use this to get the standard deviation.
answered Mar 21 at 8:12
Balakrishnan RajanBalakrishnan Rajan
1519
$endgroup$
add a comment |
$begingroup$
Every customer can order either A or B with probability 0.8 or 0.2 respectively (lets call these $p_A$ and $p_B = 1 - p_A$). The cost of A is $45 and B is $38 (call these $r_A$ and $r_B$).
Expected value of revenue from one customer (say $r/c$) is,
$$E[r/c] = (p_A times r_A) + (p_B times r_B)$$
$$E[r/c] = (p_A times r_A) + ((1-p_A) times r_B)$$
Expected revenue from 6 customers is just 6 times the expression above. This is also the mean revenue from 6 customers (let's call this $mu$).
As for the standard deviation, it is,
begin{equation}
sigma = sqrt{Var(r)}
end{equation}
where,
$$Var(r) = left(sum Pr(r = r_i) times r_i^2 right) - mu^2$$
Now for a single customer $r_i$ could only be either $r_A$ or $r_B$, but for 6 customers, there are $2^6$ different permutations of meals that can be generated for a given set of customers.
However, if $n_A$ customers chose A and $n_B = 6 - n_A$ customers chose B, then the revenue generated is $n_A times r_A + (6-n_A) times r_B$. As $n_A$ can be anywhere between 0 (no customer chose A) to 6 (all of them chose A). The probability that $n_A$ will choose A and $6-n_A$ choose B is $p_A^{n_A} (1-p_A)^{6-n_A}$.
$$Var(r) = left(sum_{n_A = 0}^{6} p_A^{n_A} (1-p_A)^{6-n_A} (n_A times r_A + (6-n_A) times r_B)^2 right) - mu^2$$
Use this to get the standard deviation.
answered Mar 21 at 8:12
Balakrishnan RajanBalakrishnan Rajan
1519
$endgroup$
add a comment |
$begingroup$
Every customer can order either A or B with probability 0.8 or 0.2 respectively (lets call these $p_A$ and $p_B = 1 - p_A$). The cost of A is $45 and B is $38 (call these $r_A$ and $r_B$).
Expected value of revenue from one customer (say $r/c$) is,
$$E[r/c] = (p_A times r_A) + (p_B times r_B)$$
$$E[r/c] = (p_A times r_A) + ((1-p_A) times r_B)$$
Expected revenue from 6 customers is just 6 times the expression above. This is also the mean revenue from 6 customers (let's call this $mu$).
As for the standard deviation, it is,
begin{equation}
sigma = sqrt{Var(r)}
end{equation}
where,
$$Var(r) = left(sum Pr(r = r_i) times r_i^2 right) - mu^2$$
Now for a single customer $r_i$ could only be either $r_A$ or $r_B$, but for 6 customers, there are $2^6$ different permutations of meals that can be generated for a given set of customers.
However, if $n_A$ customers chose A and $n_B = 6 - n_A$ customers chose B, then the revenue generated is $n_A times r_A + (6-n_A) times r_B$. As $n_A$ can be anywhere between 0 (no customer chose A) to 6 (all of them chose A). The probability that $n_A$ will choose A and $6-n_A$ choose B is $p_A^{n_A} (1-p_A)^{6-n_A}$.
$$Var(r) = left(sum_{n_A = 0}^{6} p_A^{n_A} (1-p_A)^{6-n_A} (n_A times r_A + (6-n_A) times r_B)^2 right) - mu^2$$
Use this to get the standard deviation.
answered Mar 21 at 8:12
Balakrishnan RajanBalakrishnan Rajan
1519
$endgroup$
Every customer can order either A or B with probability 0.8 or 0.2 respectively (lets call these $p_A$ and $p_B = 1 - p_A$). The cost of A is $45 and B is $38 (call these $r_A$ and $r_B$).
Expected value of revenue from one customer (say $r/c$) is,
$$E[r/c] = (p_A times r_A) + (p_B times r_B)$$
$$E[r/c] = (p_A times r_A) + ((1-p_A) times r_B)$$
Expected revenue from 6 customers is just 6 times the expression above. This is also the mean revenue from 6 customers (let's call this $mu$).
As for the standard deviation, it is,
begin{equation}
sigma = sqrt{Var(r)}
end{equation}
where,
$$Var(r) = left(sum Pr(r = r_i) times r_i^2 right) - mu^2$$
Now for a single customer $r_i$ could only be either $r_A$ or $r_B$, but for 6 customers, there are $2^6$ different permutations of meals that can be generated for a given set of customers.
However, if $n_A$ customers chose A and $n_B = 6 - n_A$ customers chose B, then the revenue generated is $n_A times r_A + (6-n_A) times r_B$. As $n_A$ can be anywhere between 0 (no customer chose A) to 6 (all of them chose A). The probability that $n_A$ will choose A and $6-n_A$ choose B is $p_A^{n_A} (1-p_A)^{6-n_A}$.
$$Var(r) = left(sum_{n_A = 0}^{6} p_A^{n_A} (1-p_A)^{6-n_A} (n_A times r_A + (6-n_A) times r_B)^2 right) - mu^2$$
Use this to get the standard deviation.
answered Mar 21 at 8:12
Balakrishnan RajanBalakrishnan Rajan
1519
edited Mar 24 at 13:30
answered Mar 21 at 8:12
Balakrishnan RajanBalakrishnan Rajan
1519
answered Mar 21 at 8:12
Balakrishnan RajanBalakrishnan Rajan
1519
answered Mar 21 at 8:12
Balakrishnan RajanBalakrishnan Rajan
1519
1519
add a comment |
add a comment |
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