Question
Park rangers have been recording the number of black bear cubs born in a national preserve over the past 100 years. According to the following table, what is the estimated probability that there will be more than 100 black bear cubs born in any given year?
Number of Cubs Born in a Year Frequency
0-25 3
26-50 6
51-75 18
76-100 32
101-125 24
126-150 10
151-175 7
(1 point)
Responses
59%
59%
34%
34%
24%
24%
41%
Number of Cubs Born in a Year Frequency
0-25 3
26-50 6
51-75 18
76-100 32
101-125 24
126-150 10
151-175 7
(1 point)
Responses
59%
59%
34%
34%
24%
24%
41%
Answers
GPT-4o mini
To determine the probability of more than 100 black bear cubs being born in a given year, we need to sum the frequencies of the ranges where the number of cubs born exceeds 100. According to the table:
- 101-125: 24
- 126-150: 10
- 151-175: 7
Now, let's calculate the total frequency of cubs born over 100:
\[
24 + 10 + 7 = 41
\]
Next, we need to find the total number of frequency counts from the entire table:
\[
3 + 6 + 18 + 32 + 24 + 10 + 7 = 100
\]
Now we can calculate the probability of having more than 100 cubs:
\[
\text{Probability} = \frac{\text{Number of Years with more than 100 Cubs}}{\text{Total Number of Years}} = \frac{41}{100} = 0.41
\]
Converting this to a percentage:
\[
0.41 \times 100 = 41\%
\]
Thus, the estimated probability that there will be more than 100 black bear cubs born in any given year is **41%**.
- 101-125: 24
- 126-150: 10
- 151-175: 7
Now, let's calculate the total frequency of cubs born over 100:
\[
24 + 10 + 7 = 41
\]
Next, we need to find the total number of frequency counts from the entire table:
\[
3 + 6 + 18 + 32 + 24 + 10 + 7 = 100
\]
Now we can calculate the probability of having more than 100 cubs:
\[
\text{Probability} = \frac{\text{Number of Years with more than 100 Cubs}}{\text{Total Number of Years}} = \frac{41}{100} = 0.41
\]
Converting this to a percentage:
\[
0.41 \times 100 = 41\%
\]
Thus, the estimated probability that there will be more than 100 black bear cubs born in any given year is **41%**.