Answer :
a) The number of students who ran the mile in less than 6 minutes are 802 students.
b) The time taken for the slowest 10% of the students to run the mile is 6.16132 minutes
c) The percentage of students who ran the mile in between 400 and 500 seconds is 67.62 %.
Given data:
a. To find the number of students who ran the mile in less than 6 minutes, we need to calculate the z-score for 6 minutes and then use the standard normal distribution table (or a calculator).
First, calculate the z-score:
z = (x - μ) / σ
z = (6 - 7.11) / 0.74
z ≈ -1.5
Now, using the standard normal distribution table or a calculator, find the cumulative probability for z = -1.5. This gives us the proportion of students who ran the mile in less than 6 minutes.
Let's denote this cumulative probability as P(Z < -1.5), which represents the proportion of students who ran the mile in less than 6 minutes. To find the approximate number of students, we multiply this proportion by the total number of students (12,000):
The value of P ( z < -1.5 ) = 0.066807
Number of students = P(Z < -1.5) * 12,000
So, n = 0.066807 * 12,000
n = 801.684
n ≈ 802 students
b.
To find the time it took for the slowest 10% of the students to run the mile, we need to find the z-score corresponding to the 10th percentile and then convert it back to minutes using the mean and standard deviation.
First, find the z-score for the 10th percentile:
z(0.10) = invNorm(0.10)
= -1.282
Now, convert the z-score back to minutes:
x = μ + z * σ
Time = 7.11 + ( -1.282 ) * 0.74
Time = 6.16132 minutes
c.
To estimate the percent of students who ran the mile in between 400 and 500 seconds, we need to convert these values to z-scores and then find the cumulative probability between these z-scores.
First, convert 400 and 500 seconds to z-scores:
z1 = (400 - 7.11 * 60) / (0.74 * 60)
z1 = -0.599
z2 = (500 - 7.11 * 60) / (0.74 * 60)
z2 = 1.65315
Now, find the cumulative probability between these z-scores:
P(z1 < Z < z2)
P ( -0.599 < z < 1.65315 ) = 0.67626
Hence, the probability is 0.67626 and the percentage is 67.62%.
To learn more about z-score, refer:
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Final answer:
This answer explains how a Normal Distribution is used to calculate probabilities and estimates based on mean and standard deviation. The solution involves converting time to z-scores to derive the estimated number of students achieving certain run times.
Explanation:
This problem involves the use of the standard Normal Distribution in statistics. The first part of the question, to estimate how many students ran the mile in less than 6 minutes is done by calculating the z-score, which indicates how many standard deviations an element is from the mean. To calculate the z-score, we subtract the desired time (6 minutes) from the mean (7.11 minutes) and divide by the standard deviation (0.74 minutes). The z-score is -1.5 approximately. Looking this up in a z-table, we find the area to the left of this z-score is approximately 0.0668, so about 6.68% of the 12,000 students, or approximately 800 students, ran the mile in less than 6 minutes.
The second part of the question wants to know the time it took the slowest 10% of the students to run the mile. We look up the z-score corresponding to the leftmost 10% in a z-table, which is approximately -1.28. We can then solve for the time by plugging into the z-score formula and obtaining time = mean + z*standard deviation = 7.11 min - 1.28*0.74 min = 6.16 minutes.
Finally, converting minutes to seconds involves multiplying by 60. A mile time of 400 to 500 seconds is equivalent to 6.67 to 8.33 minutes. To estimate the percent of students who ran the mile in this span, one would find the z-scores for these times, look up the proportion of students corresponding to those z-scores using a z-table or calculator, and subtract to find the percentage that is between those two times.
Learn more about Normal Distribution here:
https://brainly.com/question/34741155
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