Probability

[Steve123]

There are 200 participants in a case study. They are tasting one of two soft drinks to see if they can distinguish which one they are drinking. They have no ability to distinguish the two so equal chance. How would I work out the probability of .5 to .6 picking the correct drink? How could I work this out in Excel?

 

[BigBeachBanana]

There are 200 participants in a case study. They are tasting one of two soft drinks to see if they can distinguish which one they are drinking. They have no ability to distinguish the two so equal chance. How would I work out the probability of .5 to .6 picking the correct drink? How could I work this out in Excel?

Is this a homework question or something else? Also, what do you know about binomial distribution?

 

[Steven G]

This is a math help forum. As such. no one here is going to solve your problem for you. You need to show us what you have done so far sot we can help guide you to your solution.

 

[Deleted member 4993]

There are 200 participants in a case study. They are tasting one of two soft drinks to see if they can distinguish which one they are drinking. They have no ability to distinguish the two so equal chance. How would I work out the probability of .5 to .6 picking the correct drink? How could I work this out in Excel?

How do you define "correct drink"?

Please post the exact problem as it was given to you.

 

[Dr.Peterson]

There are 200 participants in a case study. They are tasting one of two soft drinks to see if they can distinguish which one they are drinking. They have no ability to distinguish the two so equal chance. How would I work out the probability of .5 to .6 picking the correct drink? How could I work this out in Excel?

This could be written more clearly. As I understand it, there are 200 people, and each essentially guesses randomly which of two drinks they are given, so that each has a 50% chance of guessing correctly. You are asked for, I think, the probability that the percentage of correct answers is between 50% and 60%.

As for using Excel, you can do that only if you know what method is required. You've been given a hint: the binomial distribution will be appropriate.

Tell us what you want Excel to do for you, and we can help you use it. But you first need to learn how to determine that. (We're assuming this is for a class.)

 

[Steve123]

This could be written more clearly. As I understand it, there are 200 people, and each essentially guesses randomly which of two drinks they are given, so that each has a 50% chance of guessing correctly. You are asked for, I think, the probability that the percentage of correct answers is between 50% and 60%.

As for using Excel, you can do that only if you know what method is required. You've been given a hint: the binomial distribution will be appropriate.

Tell us what you want Excel to do for you, and we can help you use it. But you first need to learn how to determine that. (We're assuming this is for a class.)

It's for an exam revision. I have been given the answer of .4977 but the calculations were just from a PHStat screenshot. I'm after an equation so I know for future questions of similarity. I've used BINORM.DIST on .5 of 200 and .6 of 200 then worked out the difference but my answer was .470.

 

[Dr.Peterson]

It's for an exam revision. I have been given the answer of .4977 but the calculations were just from a PHStat screenshot. I'm after an equation so I know for future questions of similarity. I've used BINORM.DIST on .5 of 200 and .6 of 200 then worked out the difference but my answer was .470.

Please show the actual contents of those cells so we can check the details (and verify your spelling!). Also, please provide an image of the problem, and perhaps also of the answer.

So, this is an exam on which you can use Excel?

 

[blamocur]

The answer depends on whether the border values are included or not: the range of [0.5, 0.6] corresponds to [100,120] answers. My NumPy script shows :
0.5253316615062532 for [100, 119]
0.5263567655283634 for [100, 120]
0.4689831824969967 for [101, 119]
0.4700082865191069 for [101, 120]

where the ranges in the square brackets are inclusive. If we average all-inclusive [100,120] and all-exclusive [101,119] case we get 0.4976699740126801, which is probably where your given answer comes from.

 

[AvgStudent]

The answer depends on whether the border values are included or not: the range of [0.5, 0.6] corresponds to [100,120] answers. My NumPy script shows :
0.5253316615062532 for [100, 119]
0.5263567655283634 for [100, 120]
0.4689831824969967 for [101, 119]
0.4700082865191069 for [101, 120]

where the ranges in the square brackets are inclusive. If we average all-inclusive [100,120] and all-exclusive [101,119] case we get 0.4976699740126801, which is probably where your given answer comes from.

That's weird. Excel tells me that 120 corresponds to .998, whereas 101 corresponds to .583 and 102 corresponds to .638 ?

 

[blamocur]

That's weird. Excel tells me that 120 corresponds to .998, whereas 101 corresponds to .583 and 102 corresponds to .638 ?

Not strange really: your Excell will probably tell you that 100 corresponds to 0.5282, from which you would get [imath]p_{120} - p_{100} \approx0.470[/imath].

 

[AvgStudent]

Not strange really: your Excell will probably tell you that 100 corresponds to 0.5282, from which you would get [imath]p_{120} - p_{100} \approx0.470[/imath].

You're correct, but I'm unsure about how you find the limits [imath][0.5,0.6] \implies [100,120]?[/imath]

 

[blamocur]

You're correct, but I'm unsure about how you find the limits [imath][0.5,0.6] \implies [100,120]?[/imath]

By multiplying by 200.

 

[Dr.Peterson]

It's for an exam revision. I have been given the answer of .4977 but the calculations were just from a PHStat screenshot. I'm after an equation so I know for future questions of similarity. I've used BINORM.DIST on .5 of 200 and .6 of 200 then worked out the difference but my answer was .470.

The answer depends on whether the border values are included or not: the range of [0.5, 0.6] corresponds to [100,120] answers. My NumPy script shows :
0.5253316615062532 for [100, 119]
0.5263567655283634 for [100, 120]
0.4689831824969967 for [101, 119]
0.4700082865191069 for [101, 120]

where the ranges in the square brackets are inclusive. If we average all-inclusive [100,120] and all-exclusive [101,119] case we get 0.4976699740126801, which is probably where your given answer comes from.

I suspect they might be using the normal distribution without a continuity correction.

That sort of fits the unclear specification of "the probability of .5 to .6"; they aren't concerned about inclusion, because they are seeing the numbers as points on a smooth curve.

Now I want even more to see the wording of the actual problem.

You're correct, but I'm unsure about how you find the limits [imath][0.5,0.6] \implies [100,120]?[/imath]

The numbers .5 and .6 presumably mean 50% and 60% of the 200; 50% of 200 is 100, and 60% of 200 is 120.

 

[AvgStudent]

By multiplying by 200.

Got it. I think I misunderstood the question.

I suspect they might be using the normal distribution without a continuity correction.

I think you're correct. If we use the normal approximation [imath]X \sim N(100,50)[/imath].
\(\displaystyle P\left(Z\le \frac{120-100}{\sqrt{50}}\right)-P\left(Z\le \frac{100-100}{\sqrt{50}}\right)=0.9977-0.5=0.4977\)