Probability of hitting a target

[Agent Smith]

TL;DR Summary

Expected number of shots on target

Say an archer has a probability p of hitting the target.
Given n shots at the target, the number of hits = np
The standard deviation of hits = $\sqrt{np(1-p)}$

Say p = 0.7
Given 100 shots, my expected/average number of hits = $100 \times 0.7 = 70$
The standard deviation for the number of hits = $\sqrt {100 \times 0.7 \times 0.3} \approx 6$

We can assume this to be a normal distribution because the number of successes > 10 and the number of failures > 10.

Does this mean that ($2 \times 6 = 12$) 95% of the time the average number of hits = $70 \pm 12$ (mean $\pm 2 \times$ standard deviation)?

So we have an interval (58, 82). Is this a confidence interval or something else?

 

[Hornbein]

TL;DR Summary: Expected number of shots on target

Say an archer has a probability p of hitting the target.
Given n shots at the target, the number of hits = np
The standard deviation of hits = $\sqrt{np(1-p)}$

Say p = 0.7
Given 100 shots, my expected/average number of hits = $100 \times 0.7 = 70$
The standard deviation for the number of hits = $\sqrt {10 \times 0.7 \times 0.3} \approx 1.4$

We can assume this to be a normal distribution because the number of successes > 10 and the number of failures > 10.

Does this mean that ($2 \times 1.4 = 2.8$) 95% of the time the average number of hits = $70 \pm 2.8$ (mean $\pm 2 \times$ standard deviation)?

That standard deviation is too small. You dropped a zero. n = 100, not 10.

 

[Agent Smith]

That standard deviation is too small. You dropped a zero. n = 100, not 10.

I made some edits.

 

[Hornbein]

Looks OK to me.

Take out "average." If we did an actual trial we'd just count the number of hits and wouldn't be averaging anything.

On average in 95% of trials the number of hits is in the interval 70±12. That is, it isn't guaranteed to be 95%. The actual percentage of 100-shot trials that would meet this criterion is a random variable with expected value of 95%.

 

[Agent Smith]

Looks OK to me.

Take out "average." If we did an actual trial we'd just count the number of hits and wouldn't be averaging anything.

On average in 95% of trials the number of hits is in the interval 70±12. That is, it isn't guaranteed to be 95%. The actual percentage of trials that would meet this criterion is a random variable with expected value of 95%.

That is to say the number of hits will be in the interval (58, 82), 95% of the time?

 

[FactChecker]

Does this mean that ($2 \times 6 = 12$) 95% of the time the average number of hits = $70 \pm 12$

You should say "is within $\pm 12$ of $70$", not "=".

(mean $\pm 2 \times$ standard deviation)?

So we have an interval (58, 82). Is this a confidence interval or something else?

It is the 95% confidence interval. There are other intervals with different confidence levels.

 

[Hornbein]

That is to say the number of hits will be in the interval (58, 82), 95% of the time?

That's the expectation. In real life it might be different. In math we might say that it converges to 95% as the number of 100-shot trials goes to infinity.

 

[Agent Smith]

@Hornbein can I find the probability of hitting the target 64/less times in 100 attempts now?

$\text{z score} = \frac{ 64- 70}{6} = -1$
The probability associated with a $\text{z score} = -1$ is $0.16$
$P(X \leq 64) = 0.16$

 

[Hornbein]

@Hornbein can I find the probability of hitting the target 64/less times in 100 attempts now?

$\text{z score} = \frac{ 64- 70}{6} = -1$
The probability associated with a $\text{z score} = -1$ is $0.16$
$P(X \leq 64) = 0.16$

Looks good to me.

 

[Agent Smith]

Looks good to me.

Are you sure?

 

[Hornbein]

No. I don't get paid to do this.