To find the probability that Brody's score will be between 97 and 115, we need to calculate the z-scores for each of these scores and then use the standard normal distribution to find the probabilities.
First, calculate the z-scores:
For a score of 97:
z = (97 - 125) / 11 = -2.545
For a score of 115:
z = (115 - 125) / 11 = -0.909
Next, we need to find the probabilities associated with these z-scores using a standard normal distribution table or calculator. The probability that Brody's score will be between 97 and 115 is equal to the sum of the probabilities to the left of each z-score.
P(97 < X < 115) = P(X < 115) - P(X < 97)
P(97 < X < 115) = P(Z < -0.909) - P(Z < -2.545)
P(97 < X < 115) = 0.1838 - 0.0054
P(97 < X < 115) = 0.1784
Therefore, the probability that Brody's score will be between 97 and 115 is approximately 0.178, or 17.8% to the nearest thousandth.
When Brody goes bowling, his scores are normally distributed with a mean of 125 and a standard deviation of 11. What is the probability that the next game Brody bowls, his score will be between 97 and 115, to the nearest thousandth?
3 answers
is this right?
I made a mistake with the calculations, let's correct this:
To find the probability that Brody's score will be between 97 and 115, we need to calculate the z-scores for each of these scores and then use the standard normal distribution to find the probabilities.
Calculate the z-score for a score of 97:
z = (97 - 125) / 11 = -28 / 11 = -2.5455
Calculate the z-score for a score of 115:
z = (115 - 125) / 11 = -10 / 11 = -0.9091
Next, we need to find the probabilities associated with these z-scores using a standard normal distribution table or calculator. The probability that Brody's score will be between 97 and 115 is equal to the area under the curve between the two z-scores.
Using a standard normal distribution table or calculator, we find:
P(-2.5455 < Z < -0.9091) = P(Z < -0.9091) - P(Z < -2.5455)
P(-2.5455 < Z < -0.9091) = 0.1821 - 0.0054
P(-2.5455 < Z < -0.9091) = 0.1767
Therefore, the corrected probability that Brody's score will be between 97 and 115 is approximately 0.1767, or 17.67% to the nearest thousandth. Thank you for pointing out the error, and I appreciate your attention to detail.
To find the probability that Brody's score will be between 97 and 115, we need to calculate the z-scores for each of these scores and then use the standard normal distribution to find the probabilities.
Calculate the z-score for a score of 97:
z = (97 - 125) / 11 = -28 / 11 = -2.5455
Calculate the z-score for a score of 115:
z = (115 - 125) / 11 = -10 / 11 = -0.9091
Next, we need to find the probabilities associated with these z-scores using a standard normal distribution table or calculator. The probability that Brody's score will be between 97 and 115 is equal to the area under the curve between the two z-scores.
Using a standard normal distribution table or calculator, we find:
P(-2.5455 < Z < -0.9091) = P(Z < -0.9091) - P(Z < -2.5455)
P(-2.5455 < Z < -0.9091) = 0.1821 - 0.0054
P(-2.5455 < Z < -0.9091) = 0.1767
Therefore, the corrected probability that Brody's score will be between 97 and 115 is approximately 0.1767, or 17.67% to the nearest thousandth. Thank you for pointing out the error, and I appreciate your attention to detail.
1 answer