Carter is participating in a mousetrap car competition. The goal is to have the mousetrap car that travels the farthest. Carter records the distance for each competitor, including his own, and finds the data approximately follow a normal distribution with a mean of 8.4 meters and a standard deviation of 2.3 meters. Carter's car went 10.47 meters, and he believes that was farther than 90%
of his competitors. Explain whether Carter is correct
To determine whether Carter's car went farther than 90% of his competitors, we need to find the z-score corresponding to the 90th percentile and see if the z-score for Carter's car is greater than this value.
First, we calculate the z-score for Carter's car using the formula:
z = (x - μ) / σ
Where x is the value (distance) for Carter's car, μ is the mean distance, and σ is the standard deviation.
Plugging in the given values:
z = (10.47 - 8.4) / 2.3
z = 0.913
Next, we find the z-score corresponding to the 90th percentile using a z-score table or calculator. The z-score for the 90th percentile is approximately 1.282.
Since Carter's z-score of 0.913 is less than the z-score for the 90th percentile of 1.282, it means that his car did not go farther than 90% of his competitors. Therefore, Carter's belief is incorrect.
A) Carter is not correct. A z
-score of 0.9
corresponds to an area of 0.8159 under the normal curve and to the left of 10.47
. His car went farther than approximately 81.59% of competitors' cars.
B) Carter is not correct. A z
-score of 0.9
corresponds to an area of 0.1841 under the normal curve and to the right of 10.47
. His car went farther than approximately 18.41% of competitors' cars.
C) Carter is correct. A z
-score of 2.07 corresponds to an area of 0.9808 under the normal curve and to the left of 10.47
. His car went farther than approximately 98.08% of competitors' cars.
C)Carter is correct. A z
-score of 0.9
corresponds to an area of 0.9
under the normal curve and to the left of 10.47
. His car went farther than approximately 90%
of competitors' cars.