Approximation of exponential and normal probabilities
$begingroup$
A company uses a portable high-intensity flashlight. Batteries and bulbs burn out quickly.
The lifetime of batteries has Exponential distribution with mean 10 hours. The bulbs have
lifetimes that are Normally distributed with mean 32 and standard deviation 5. Assume batteries
and bulbs are randomly sampled.
Find the probabilities for the following events. Where appropriate you may approximate
probabilities.
a) A battery lasts over 11 hours.
b) A sample of 20 batteries has a sample mean over 11 hours.
c) A sample of 200 batteries has a sample mean over 11 hours.
I'm not sure how to attack these questions because I've only learned approximating with the normal distribution to the binomial, any suggestions?
statistics
$endgroup$
|
show 1 more comment
$begingroup$
A company uses a portable high-intensity flashlight. Batteries and bulbs burn out quickly.
The lifetime of batteries has Exponential distribution with mean 10 hours. The bulbs have
lifetimes that are Normally distributed with mean 32 and standard deviation 5. Assume batteries
and bulbs are randomly sampled.
Find the probabilities for the following events. Where appropriate you may approximate
probabilities.
a) A battery lasts over 11 hours.
b) A sample of 20 batteries has a sample mean over 11 hours.
c) A sample of 200 batteries has a sample mean over 11 hours.
I'm not sure how to attack these questions because I've only learned approximating with the normal distribution to the binomial, any suggestions?
statistics
$endgroup$
1
$begingroup$
I got a 0 probability for a) so I think there must've been something wrong in the calculation there, I put in 1-pexp(11,10) into R. However, with b) and c) I'm not sure how to solve it with the sample numbers in there.
$endgroup$
– J.Doe
Feb 20 '17 at 3:55
$begingroup$
In R functionpexp
the second argument is the rate, not the mean. So use1/10
. // Since you're using R, I'll mention that sums and averages of exponential samples have gamma distributions; have you studied that? In R, the 3rd argument ofpgamma
is also rate $lambda = 1/beta,$ where $beta$ is scale parameter.
$endgroup$
– BruceET
Feb 20 '17 at 4:00
1
$begingroup$
Wow I can't believe I forgot about that thanks, haven't taken stats in a little bit as you can probably tell.
$endgroup$
– J.Doe
Feb 20 '17 at 4:03
$begingroup$
Suggest you look at this. Including the answer and the Wikipedia ref.
$endgroup$
– BruceET
Feb 20 '17 at 4:15
$begingroup$
Sorry, there's an error in that link, see my Answer to (b). I have posted a correction note at the link. Please doublecheck.
$endgroup$
– BruceET
Feb 20 '17 at 17:30
|
show 1 more comment
$begingroup$
A company uses a portable high-intensity flashlight. Batteries and bulbs burn out quickly.
The lifetime of batteries has Exponential distribution with mean 10 hours. The bulbs have
lifetimes that are Normally distributed with mean 32 and standard deviation 5. Assume batteries
and bulbs are randomly sampled.
Find the probabilities for the following events. Where appropriate you may approximate
probabilities.
a) A battery lasts over 11 hours.
b) A sample of 20 batteries has a sample mean over 11 hours.
c) A sample of 200 batteries has a sample mean over 11 hours.
I'm not sure how to attack these questions because I've only learned approximating with the normal distribution to the binomial, any suggestions?
statistics
$endgroup$
A company uses a portable high-intensity flashlight. Batteries and bulbs burn out quickly.
The lifetime of batteries has Exponential distribution with mean 10 hours. The bulbs have
lifetimes that are Normally distributed with mean 32 and standard deviation 5. Assume batteries
and bulbs are randomly sampled.
Find the probabilities for the following events. Where appropriate you may approximate
probabilities.
a) A battery lasts over 11 hours.
b) A sample of 20 batteries has a sample mean over 11 hours.
c) A sample of 200 batteries has a sample mean over 11 hours.
I'm not sure how to attack these questions because I've only learned approximating with the normal distribution to the binomial, any suggestions?
statistics
statistics
asked Feb 20 '17 at 2:49
J.DoeJ.Doe
226
226
1
$begingroup$
I got a 0 probability for a) so I think there must've been something wrong in the calculation there, I put in 1-pexp(11,10) into R. However, with b) and c) I'm not sure how to solve it with the sample numbers in there.
$endgroup$
– J.Doe
Feb 20 '17 at 3:55
$begingroup$
In R functionpexp
the second argument is the rate, not the mean. So use1/10
. // Since you're using R, I'll mention that sums and averages of exponential samples have gamma distributions; have you studied that? In R, the 3rd argument ofpgamma
is also rate $lambda = 1/beta,$ where $beta$ is scale parameter.
$endgroup$
– BruceET
Feb 20 '17 at 4:00
1
$begingroup$
Wow I can't believe I forgot about that thanks, haven't taken stats in a little bit as you can probably tell.
$endgroup$
– J.Doe
Feb 20 '17 at 4:03
$begingroup$
Suggest you look at this. Including the answer and the Wikipedia ref.
$endgroup$
– BruceET
Feb 20 '17 at 4:15
$begingroup$
Sorry, there's an error in that link, see my Answer to (b). I have posted a correction note at the link. Please doublecheck.
$endgroup$
– BruceET
Feb 20 '17 at 17:30
|
show 1 more comment
1
$begingroup$
I got a 0 probability for a) so I think there must've been something wrong in the calculation there, I put in 1-pexp(11,10) into R. However, with b) and c) I'm not sure how to solve it with the sample numbers in there.
$endgroup$
– J.Doe
Feb 20 '17 at 3:55
$begingroup$
In R functionpexp
the second argument is the rate, not the mean. So use1/10
. // Since you're using R, I'll mention that sums and averages of exponential samples have gamma distributions; have you studied that? In R, the 3rd argument ofpgamma
is also rate $lambda = 1/beta,$ where $beta$ is scale parameter.
$endgroup$
– BruceET
Feb 20 '17 at 4:00
1
$begingroup$
Wow I can't believe I forgot about that thanks, haven't taken stats in a little bit as you can probably tell.
$endgroup$
– J.Doe
Feb 20 '17 at 4:03
$begingroup$
Suggest you look at this. Including the answer and the Wikipedia ref.
$endgroup$
– BruceET
Feb 20 '17 at 4:15
$begingroup$
Sorry, there's an error in that link, see my Answer to (b). I have posted a correction note at the link. Please doublecheck.
$endgroup$
– BruceET
Feb 20 '17 at 17:30
1
1
$begingroup$
I got a 0 probability for a) so I think there must've been something wrong in the calculation there, I put in 1-pexp(11,10) into R. However, with b) and c) I'm not sure how to solve it with the sample numbers in there.
$endgroup$
– J.Doe
Feb 20 '17 at 3:55
$begingroup$
I got a 0 probability for a) so I think there must've been something wrong in the calculation there, I put in 1-pexp(11,10) into R. However, with b) and c) I'm not sure how to solve it with the sample numbers in there.
$endgroup$
– J.Doe
Feb 20 '17 at 3:55
$begingroup$
In R function
pexp
the second argument is the rate, not the mean. So use 1/10
. // Since you're using R, I'll mention that sums and averages of exponential samples have gamma distributions; have you studied that? In R, the 3rd argument of pgamma
is also rate $lambda = 1/beta,$ where $beta$ is scale parameter.$endgroup$
– BruceET
Feb 20 '17 at 4:00
$begingroup$
In R function
pexp
the second argument is the rate, not the mean. So use 1/10
. // Since you're using R, I'll mention that sums and averages of exponential samples have gamma distributions; have you studied that? In R, the 3rd argument of pgamma
is also rate $lambda = 1/beta,$ where $beta$ is scale parameter.$endgroup$
– BruceET
Feb 20 '17 at 4:00
1
1
$begingroup$
Wow I can't believe I forgot about that thanks, haven't taken stats in a little bit as you can probably tell.
$endgroup$
– J.Doe
Feb 20 '17 at 4:03
$begingroup$
Wow I can't believe I forgot about that thanks, haven't taken stats in a little bit as you can probably tell.
$endgroup$
– J.Doe
Feb 20 '17 at 4:03
$begingroup$
Suggest you look at this. Including the answer and the Wikipedia ref.
$endgroup$
– BruceET
Feb 20 '17 at 4:15
$begingroup$
Suggest you look at this. Including the answer and the Wikipedia ref.
$endgroup$
– BruceET
Feb 20 '17 at 4:15
$begingroup$
Sorry, there's an error in that link, see my Answer to (b). I have posted a correction note at the link. Please doublecheck.
$endgroup$
– BruceET
Feb 20 '17 at 17:30
$begingroup$
Sorry, there's an error in that link, see my Answer to (b). I have posted a correction note at the link. Please doublecheck.
$endgroup$
– BruceET
Feb 20 '17 at 17:30
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Because the link that I provided in a Comment has incorrect information,
I am posting this Answer to (b).
You have $n = 20$ observations from $mathsf{Exp}(lambda = 1/10),$
and you seek $P(bar X > 11).$
The individual observations have $mu =E(X_i) = 1/lambda = 10,$
variance $sigma^2 = 1/lambda^2 = 100,$ and SD $sigma = 1/lambda.$
The mean of $n = 20$ such observations has $bar X sim mathsf{Gamma}(n,nlambda).$ Hence $E(bar X) = frac{n}{nlambda} = frac{1}{lambda} = 10,$
variance $V(bar X) = frac{1}{nlambda^2} = frac{sigma^2}{n} =
frac{100}{20} = 5,$
and $SD(bar X) = frac{1}{lambdasqrt{n}} = frac{sigma}{sqrt{n}} =
frac{10}{sqrt{20}} =
sqrt{5} = 2.2361.$
These results can be derived using moment generating functions and
standard formulas for means and variances of random variables.
Hence $P(bar X > 11) = 0.306027,$ as computed in R statistical software below:
n = 20; lam = 0.1; 1 - pgamma(11, n, n*lam)
## 0.306027
Of course, part (c) can be done similarly in R. Also, in (c) a normal approximation
based on $n=200$ is reasonably accurate.
n = 200; lam = .1; mu = 1/lam; sg = 1/(lam*sqrt(n))
1 - pgamma(11, n, n*lam)
## 0.08180569 # exact
1 - pnorm(11, mu, sg)
## 0.0786496 # norm aprx
Note: Just as a 'reality check', when claiming an error elsewhere, I took a
million samples of size $n = 20$ from $mathsf{Exp}(rate = lambda = 0.1)$
and found the corresponding million averages with the following results,
agreeing with the theoretical results for (b) above, within the margin of sampling error.
a = replicate(10^6, mean(rexp(20, .1)))
mean(a > 11)
## 0.306304 # aprx P(Avg > 11) = 0.306027
mean(a); sd(a)
## 9.99888 # aprx 10
## 2.236212 # aprx sqrt(5)
Below is a histogram of the one million sample means along with the
density function of $mathsf{Gamma}(20, 2).$
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2152349%2fapproximation-of-exponential-and-normal-probabilities%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Because the link that I provided in a Comment has incorrect information,
I am posting this Answer to (b).
You have $n = 20$ observations from $mathsf{Exp}(lambda = 1/10),$
and you seek $P(bar X > 11).$
The individual observations have $mu =E(X_i) = 1/lambda = 10,$
variance $sigma^2 = 1/lambda^2 = 100,$ and SD $sigma = 1/lambda.$
The mean of $n = 20$ such observations has $bar X sim mathsf{Gamma}(n,nlambda).$ Hence $E(bar X) = frac{n}{nlambda} = frac{1}{lambda} = 10,$
variance $V(bar X) = frac{1}{nlambda^2} = frac{sigma^2}{n} =
frac{100}{20} = 5,$
and $SD(bar X) = frac{1}{lambdasqrt{n}} = frac{sigma}{sqrt{n}} =
frac{10}{sqrt{20}} =
sqrt{5} = 2.2361.$
These results can be derived using moment generating functions and
standard formulas for means and variances of random variables.
Hence $P(bar X > 11) = 0.306027,$ as computed in R statistical software below:
n = 20; lam = 0.1; 1 - pgamma(11, n, n*lam)
## 0.306027
Of course, part (c) can be done similarly in R. Also, in (c) a normal approximation
based on $n=200$ is reasonably accurate.
n = 200; lam = .1; mu = 1/lam; sg = 1/(lam*sqrt(n))
1 - pgamma(11, n, n*lam)
## 0.08180569 # exact
1 - pnorm(11, mu, sg)
## 0.0786496 # norm aprx
Note: Just as a 'reality check', when claiming an error elsewhere, I took a
million samples of size $n = 20$ from $mathsf{Exp}(rate = lambda = 0.1)$
and found the corresponding million averages with the following results,
agreeing with the theoretical results for (b) above, within the margin of sampling error.
a = replicate(10^6, mean(rexp(20, .1)))
mean(a > 11)
## 0.306304 # aprx P(Avg > 11) = 0.306027
mean(a); sd(a)
## 9.99888 # aprx 10
## 2.236212 # aprx sqrt(5)
Below is a histogram of the one million sample means along with the
density function of $mathsf{Gamma}(20, 2).$
$endgroup$
add a comment |
$begingroup$
Because the link that I provided in a Comment has incorrect information,
I am posting this Answer to (b).
You have $n = 20$ observations from $mathsf{Exp}(lambda = 1/10),$
and you seek $P(bar X > 11).$
The individual observations have $mu =E(X_i) = 1/lambda = 10,$
variance $sigma^2 = 1/lambda^2 = 100,$ and SD $sigma = 1/lambda.$
The mean of $n = 20$ such observations has $bar X sim mathsf{Gamma}(n,nlambda).$ Hence $E(bar X) = frac{n}{nlambda} = frac{1}{lambda} = 10,$
variance $V(bar X) = frac{1}{nlambda^2} = frac{sigma^2}{n} =
frac{100}{20} = 5,$
and $SD(bar X) = frac{1}{lambdasqrt{n}} = frac{sigma}{sqrt{n}} =
frac{10}{sqrt{20}} =
sqrt{5} = 2.2361.$
These results can be derived using moment generating functions and
standard formulas for means and variances of random variables.
Hence $P(bar X > 11) = 0.306027,$ as computed in R statistical software below:
n = 20; lam = 0.1; 1 - pgamma(11, n, n*lam)
## 0.306027
Of course, part (c) can be done similarly in R. Also, in (c) a normal approximation
based on $n=200$ is reasonably accurate.
n = 200; lam = .1; mu = 1/lam; sg = 1/(lam*sqrt(n))
1 - pgamma(11, n, n*lam)
## 0.08180569 # exact
1 - pnorm(11, mu, sg)
## 0.0786496 # norm aprx
Note: Just as a 'reality check', when claiming an error elsewhere, I took a
million samples of size $n = 20$ from $mathsf{Exp}(rate = lambda = 0.1)$
and found the corresponding million averages with the following results,
agreeing with the theoretical results for (b) above, within the margin of sampling error.
a = replicate(10^6, mean(rexp(20, .1)))
mean(a > 11)
## 0.306304 # aprx P(Avg > 11) = 0.306027
mean(a); sd(a)
## 9.99888 # aprx 10
## 2.236212 # aprx sqrt(5)
Below is a histogram of the one million sample means along with the
density function of $mathsf{Gamma}(20, 2).$
$endgroup$
add a comment |
$begingroup$
Because the link that I provided in a Comment has incorrect information,
I am posting this Answer to (b).
You have $n = 20$ observations from $mathsf{Exp}(lambda = 1/10),$
and you seek $P(bar X > 11).$
The individual observations have $mu =E(X_i) = 1/lambda = 10,$
variance $sigma^2 = 1/lambda^2 = 100,$ and SD $sigma = 1/lambda.$
The mean of $n = 20$ such observations has $bar X sim mathsf{Gamma}(n,nlambda).$ Hence $E(bar X) = frac{n}{nlambda} = frac{1}{lambda} = 10,$
variance $V(bar X) = frac{1}{nlambda^2} = frac{sigma^2}{n} =
frac{100}{20} = 5,$
and $SD(bar X) = frac{1}{lambdasqrt{n}} = frac{sigma}{sqrt{n}} =
frac{10}{sqrt{20}} =
sqrt{5} = 2.2361.$
These results can be derived using moment generating functions and
standard formulas for means and variances of random variables.
Hence $P(bar X > 11) = 0.306027,$ as computed in R statistical software below:
n = 20; lam = 0.1; 1 - pgamma(11, n, n*lam)
## 0.306027
Of course, part (c) can be done similarly in R. Also, in (c) a normal approximation
based on $n=200$ is reasonably accurate.
n = 200; lam = .1; mu = 1/lam; sg = 1/(lam*sqrt(n))
1 - pgamma(11, n, n*lam)
## 0.08180569 # exact
1 - pnorm(11, mu, sg)
## 0.0786496 # norm aprx
Note: Just as a 'reality check', when claiming an error elsewhere, I took a
million samples of size $n = 20$ from $mathsf{Exp}(rate = lambda = 0.1)$
and found the corresponding million averages with the following results,
agreeing with the theoretical results for (b) above, within the margin of sampling error.
a = replicate(10^6, mean(rexp(20, .1)))
mean(a > 11)
## 0.306304 # aprx P(Avg > 11) = 0.306027
mean(a); sd(a)
## 9.99888 # aprx 10
## 2.236212 # aprx sqrt(5)
Below is a histogram of the one million sample means along with the
density function of $mathsf{Gamma}(20, 2).$
$endgroup$
Because the link that I provided in a Comment has incorrect information,
I am posting this Answer to (b).
You have $n = 20$ observations from $mathsf{Exp}(lambda = 1/10),$
and you seek $P(bar X > 11).$
The individual observations have $mu =E(X_i) = 1/lambda = 10,$
variance $sigma^2 = 1/lambda^2 = 100,$ and SD $sigma = 1/lambda.$
The mean of $n = 20$ such observations has $bar X sim mathsf{Gamma}(n,nlambda).$ Hence $E(bar X) = frac{n}{nlambda} = frac{1}{lambda} = 10,$
variance $V(bar X) = frac{1}{nlambda^2} = frac{sigma^2}{n} =
frac{100}{20} = 5,$
and $SD(bar X) = frac{1}{lambdasqrt{n}} = frac{sigma}{sqrt{n}} =
frac{10}{sqrt{20}} =
sqrt{5} = 2.2361.$
These results can be derived using moment generating functions and
standard formulas for means and variances of random variables.
Hence $P(bar X > 11) = 0.306027,$ as computed in R statistical software below:
n = 20; lam = 0.1; 1 - pgamma(11, n, n*lam)
## 0.306027
Of course, part (c) can be done similarly in R. Also, in (c) a normal approximation
based on $n=200$ is reasonably accurate.
n = 200; lam = .1; mu = 1/lam; sg = 1/(lam*sqrt(n))
1 - pgamma(11, n, n*lam)
## 0.08180569 # exact
1 - pnorm(11, mu, sg)
## 0.0786496 # norm aprx
Note: Just as a 'reality check', when claiming an error elsewhere, I took a
million samples of size $n = 20$ from $mathsf{Exp}(rate = lambda = 0.1)$
and found the corresponding million averages with the following results,
agreeing with the theoretical results for (b) above, within the margin of sampling error.
a = replicate(10^6, mean(rexp(20, .1)))
mean(a > 11)
## 0.306304 # aprx P(Avg > 11) = 0.306027
mean(a); sd(a)
## 9.99888 # aprx 10
## 2.236212 # aprx sqrt(5)
Below is a histogram of the one million sample means along with the
density function of $mathsf{Gamma}(20, 2).$
edited Feb 20 '17 at 22:44
answered Feb 20 '17 at 9:06
BruceETBruceET
35.3k71440
35.3k71440
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2152349%2fapproximation-of-exponential-and-normal-probabilities%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
I got a 0 probability for a) so I think there must've been something wrong in the calculation there, I put in 1-pexp(11,10) into R. However, with b) and c) I'm not sure how to solve it with the sample numbers in there.
$endgroup$
– J.Doe
Feb 20 '17 at 3:55
$begingroup$
In R function
pexp
the second argument is the rate, not the mean. So use1/10
. // Since you're using R, I'll mention that sums and averages of exponential samples have gamma distributions; have you studied that? In R, the 3rd argument ofpgamma
is also rate $lambda = 1/beta,$ where $beta$ is scale parameter.$endgroup$
– BruceET
Feb 20 '17 at 4:00
1
$begingroup$
Wow I can't believe I forgot about that thanks, haven't taken stats in a little bit as you can probably tell.
$endgroup$
– J.Doe
Feb 20 '17 at 4:03
$begingroup$
Suggest you look at this. Including the answer and the Wikipedia ref.
$endgroup$
– BruceET
Feb 20 '17 at 4:15
$begingroup$
Sorry, there's an error in that link, see my Answer to (b). I have posted a correction note at the link. Please doublecheck.
$endgroup$
– BruceET
Feb 20 '17 at 17:30