SOLUTION: Records show that the probability of seeing a hawk migrating on a day in
September is about 35%. What is the minimum number of days a person must
watch to be at least 96.8% sure
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-> SOLUTION: Records show that the probability of seeing a hawk migrating on a day in
September is about 35%. What is the minimum number of days a person must
watch to be at least 96.8% sure
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Question 1192723: Records show that the probability of seeing a hawk migrating on a day in
September is about 35%. What is the minimum number of days a person must
watch to be at least 96.8% sure of seeing one or more hawks migrating?
You can put this solution on YOUR website! To solve this, we'll use the concept of the complement of the binomial distribution.
* **Probability of not seeing a hawk on a single day:** 1 - 0.35 = 0.65
* **Probability of not seeing a hawk on 'n' consecutive days:** 0.65^n
* **Probability of seeing at least one hawk in 'n' days:** 1 - 0.65^n
We want to find the minimum 'n' such that:
1 - 0.65^n >= 0.968
0.65^n <= 0.032
Taking the natural logarithm of both sides:
n * ln(0.65) <= ln(0.032)
n >= ln(0.032) / ln(0.65)
n >= 7.85
Since 'n' must be an integer, the minimum number of days a person must watch is **8 days**.
