Statistics 101: Simple Binomial Sales Quota Analysis

[@vulnerablerummy]

09:00 for those of you who haven't know it, if you don't have a ti calculator on hand, you can still calculate this on excel (2016) with the function "=BINOM.DIST(number_s; trial; probability_s; cumulative)" where : number_s : number of 'success', which in this case is successful sales (input 1 for probability of 1 success, input 2 for probability of 2 success, etc.) trial : number of trials, 10 for Joan, 16 for Margo probability : percentage of 'success' cumulative : fill this with 'true' if you want to add probability of multiple number_s (such as 1, 2, 3, 4, and 5), or 'false' if you just want the probability of a single number_s

[@vulnerablerummy]

Mr. Brandon, thank you for liking my comment. If you read this, i still can't find conclusive explanation to your bonus questions. i'll be delighted if it's possible for you to share the explanations to us in any format, only if it doesn't burden you much. god bless edit : from 17:05 to 17:16 i guess this is the answer to my questions and many others. Mr. Brandon mentions there is no precise way to calculate it, we basically try calculating every potential number (from 16, 17, 18, to 19 and 45%, 46%,... to 52%). i'd say this is reasonable enough for an explanation.

[@kinjalvora3352]

Maybe this is kinda simple: at_least_6 for Mary = .80 at_least_6 for Joan = .92 when the prob is .80 calls are 16 when the prob needs to be .92, calls would be (16*.92) / .80 = 18.4, rounding up to give a 19 on the other way .80 - .45 ( success rate) .92 = (.45 * .92) / .80 = .5175 or close to 52%

[@empaulstube6947]

To answer the bonus question, we must first create an Output or Performance Ratio of Margo. Margo's performance is n=16, sr=.45 which gives her .802 probability for making =>6 success calls per day. Let's make an Output or Performance Ratio from this using Joan's probability of success .922. In the 1st question, the "x" or the variable that we are looking for is the number of average calls per day Margo has to make so that she would also have a probability success rate of .922. Make a Ratio of Margo: .802 divided by 16 (average calls) will equal .05 which represents Margo's success rate per calls a day. Then use .922 of Joan. .922 divided .05 will give us 18.44. It should have been 18 calls only but Brandon have rounded it to 19. Do the same with question number two. Figure it out.

[@dileep7943]

I think it's the ceil of the calculation. n = 18 is approx 0.027 probability away from 0.92 while 19 is 0.00022 away from 0.92. So 19 gives a more closer result than 18, also applying the ceil function to 51.75% will result to 52%.

[@focuskalunde1438]

Thanks 🙏 I tried a lot but finally I successfully

[@kaliniaczek]

hi Brandon - could you explain me please how you've answered to the bonus questions. thank you

[@palaksodhi8123]

Please explain the solution for the bonus question

[@brianmoyer35]

At 12:50, should the notation be p(0

[@Surya42930]

Thank you very much Brandon....Running out of words to express thanks...Love your videos

[@tarikbouchnayf]

Your courses are amazing, you make things look easier... thanks for that

[@phsaraiva19]

The classes are perfect, but it is very frustrating to not know what are the real solution to the exercises. I understand that some of the people who commented on the video have found the answer, but this is really the right way to get to them? Just apply a direct proportion from one performance to another and that's it? How can we be sure that this is the right way to solve the exercise?

[@cbandaranayakeherath4450]

Could you please do a lesson and practical example of Bayesian Probability Analysis

[@bimalshrestha2354]

what is the general formula for binomcdf

[@ragsanoor]

Brandon, I love your videos... Thank you very much ...

[@farhimouad2860]

Thank you for your usefull efforts . please inverse the ordrer of those last two videos in the playlist. this one should be before the other one on mean and standardr deviation

[@hamzamohd.zubair1709]

This is video 7. Binomial Mean and Standard Deviation is Video 8. Because you mention in Binomial Mean and Standard Deviation that we have discussed cumulative probabilities, so it gets confusing. So it will be great if you switch places of videos 7 and 8. Thanks

[@holaaaa3047]

thanks for the info!

[@odalesaylor]

I think that it is the average number of successful calls per day that should be compared not the probability of six or more successes in a day. ( average = n (p) )

[@romanvasiura6705]

Thank you for this great video lesson)

[@xeeharshit]

Couldn't get the answer for the Bonus Question. Please give a mathematical procedure to solve it. Hope you answer it soon.

[@DavidDümig]

Not quite sure why we need a calculator if we have the individual probabilities ?

[@gauravagrawal1192]

Hi Brandon, Thank you for this. I didn't had Ti calculator, but found a function BINOMDIST in excel to do the job :)

[@renaldysamuel9271]

excel is better bro

[@raghavexplore1638]

Hello Sir, How Can I do this calculation using Minitab.

[@shahanakhan5693]

how to do the calculation in a fx 350ES calculator ?

[@massivejester]

very well made and easy to understand, thank you so much!

[@vaibhav19]

Hi Brandon...nice videos...they have been really helpful...can we have the solution for the bonus questions plz...i have tried them but have not been able to get the correct ans..thank you!!

[@thingocduytran7430]

I like this video. thank you so much.

[@MrBryonbarnard]

If Joan averages 10 sales a day, why does her distribution never include sales greater than 10 per day. If 10 is her average, I assume there are days with sales >10.

[@anandchari26]

Hello Brandon, Pls do video lectures on Negative Binomial Distribution and Geometric Distribution Thanks in advance

[@7kGreen]

You can use this Python script instead of calculator: from math import factorial as f def get_combination(n, x): return f(n)/(f(n-x)*f(x)) n = 10 p = 0.75 q = 0.25 distribution = [0]*(n+1) for i in list(range(n+1)): experiment = get_combination(10, i)*(p**i)*(q**(n-i)) print(f"{i} {experiment}") distribution[i] = experiment # binomcdf functionality: from_index = 0 to_index = 5 result = 0 for i in list(range(from_index, to_index+1)): result += distribution[i] print(f"Probability from {from_index} to {to_index} is {result}")

[@shkoder123]

I'm taking statistics online classes, and I doo all my home work based on their guidance and samples, and e-book, so most times I'm doing ok, but there are times I cant figure out why I don't get it? I fallow the examples step by step and watch some videos they are available to us, but im still not understanding how they got this answer? n=1655 p=2/5 Mu = n*p = 1655*2/5 = 622 q = 1 - p = 1- 2/5 = 3/2 SD = sqr npq SD = sqr 1655*2/5*3/6 = 19.9 this is their answer I like to know where they got the answer from, because when I do the math with Ti-84 my answer is 9.76 Please help me with this?

[@vulnerablerummy]

i'm 4 years too late, but heck with that.... your last calculations on SD is wrong, your q should be 3/5 not 3/2 so then, SD = sqr 1655 x (2/5) x (3/5) = 19.9298

[@jongcheulkim7284]

Thank you.

[@prabhudaskamath1353]

Thank you..

[@sushantsanthosh2129]

Bonus Questions:- 1.) Margo's average= probability of success x number of calls = np = 0.45 x 16 = 7.2 Joans average= probability of success x number of calls = np =.75 x 10 = 7.5 How many calls should Margo make to match Joans performance( i.e match Joans average). 0.45 x (16+X)=7.5 (X is excess number of calls needed to match Joans average) X= 0.66 = 1(approximately) Therefore, the number of calls needed to match joans performance is 16+X =16+1=17 calls. 2.) Margo's average= probability of success x number of calls = np = 0.45 x 16 = 7.2 Joans average= probability of success x number of calls = np =.75 x 10 = 7.5 By how much should Margo raise her success rate to match Joans performance( i.e match Joans average). (0.45+X) x 16 = 7.5 (X is excess percentage needed to match Joans average) . X= 0.018 = 0.02 (approximately) Therefore, the success rate needed to match Joans performance is 0.45 + 0.02 = .47 or 47%

[@bidusikhadka1831]

I did the same, but this doesn't match the answer provided in the video.

[@tanisyt]

The reason that it didn't match because the base to calculate the improvement in performance was related to the probability increment in the success rate of at least 6 calls, not overall.

[@yashashgaurav4848]

@@tanisyt Depends on what you mean by performance right? But I see why you would say that - it's just after the question that asks for cumulative performances. I don't see any other way than trial and error here for getting the answer necessary. Help me with a solution if you found a solution?