Answer:
Only statements A and D are correct.
- The pointer lands on an even number more often than expected.
- The actual probability of landing on an odd number is 46%.
Step-by-step explanation:
The image of the spinner isn't shown, but the image obtained from this same question on as another brainly question has the spinner numbered from 1 to 8 inclusive, meaning that there's an equal number of even and odd numbers, 4 each. I cannot post the image as it will violate community guidelines on here and lead to question deletion.
The spinner lands on an even number 135 times out of 250 spins.
Let the probability of the spinner landing on an even number = P(E)
Let the probability that the spinner lands on an odd number = P(O)
The expected probability of landing on an even number = P(E) = (4/8) = 0.5
Similarly, the expected probability of landing on an odd number = P(O) = (4/8) = 0.5
So, examining the options one at a time.
A. The pointer lands on an even number more often than expected.
The expected probability of landing on an even number = P(E) = (4/8) = 0.5
In 250 trials, expected number of times the spinner should land on an even number = 0.5 × 250 = 125.
So, landing on an even number 135 times indicates that the pointer lands on an even number more often than expected (135 > 125)
This statement is correct.
B. The pointer lands on an even number less often than expected.
This is a direct contradiction to statement A, which we have proved to be correct, hence, statement B is not correct.
C. The expected probability of landing on an even number is 54%.
The expected probability of landing on an even number = P(E) = (4/8) = 0.5 = 50%
This statement is not correct.
D. The actual probability of landing on an odd number is 46%.
The actual probability of landing on an odd number = 1 - (135/250) (since odd and even are the only 2 sample spaces)
The actual probability of landing on an odd number = 1 - 0.54 = 0.46 = 46%
This statement is correct.
E. It is equally likely that the pointer will land on an even or odd number.
The expected probabilities are the same and points to an equal likelihood of the pointer landing an even or odd number, but, the actual probabilities (54% for an even number and 46% for an odd number), show that it is not equally likely that the pointer will land on an even or odd number.
Hence, this statement is not correct.
Hope this Helps!!!
