2)For 2000 patients, a blood-clotting time was normally distributed with a mean of 8 seconds and a standard deviation of 3 seconds. What percent had blood-clotting times between 5 and 11 seconds?

Question

2)For 2000 patients, a blood-clotting time was normally distributed with a mean of 8 seconds and a standard deviation of 3 seconds. What percent had blood-clotting times between 5 and 11 seconds?

A)68%
B)34%
C)49/5%
D)47/5%
I chose A

4)A music teacher wants to determine the music performances of students. A survey of which group would produce a random sample?
A)students in the school band
B)students attendeing the annual jazz concert
C)students in every odd-numbered homeroom
D)every other player on the baseball roster
is it A?

Answer 1

2) A is correct

4): I think the point of this question is that the teacher wants a RANDOM sample of students. The only choice that would really give her a random sample would be C. All others are biased.

Answer 2

POITS A,B, AND C ARE COLLINEAR .AB=7CM BC =13CM WHATIS AC ANSWER

Answer 3

2) To find the percentage of patients with blood-clotting times between 5 and 11 seconds, you can use the Z-score formula. The Z-score measures how many standard deviations a data point is from the mean of a distribution.

First, calculate the Z-score for the lower bound of 5 seconds:
Z = (5 - 8) / 3 = -1

Then, calculate the Z-score for the upper bound of 11 seconds:
Z = (11 - 8) / 3 = 1

Next, use a Z-table or a Z-score calculator to find the area under the normal distribution curve between these two Z-scores. The area represents the percentage of data falling between the two values.

Looking up the Z-scores in the table, you will find that the area to the left of -1 is approximately 0.1587 and the area to the left of 1 is approximately 0.8413.

To find the percentage between these two values, subtract the smaller area from the larger area:
0.8413 - 0.1587 = 0.6826

Finally, multiply the result by 100 to get the percentage:
0.6826 * 100 = 68.26%

So the correct answer is A) 68%.

4) A random sample is one in which each member of the population has an equal chance of being selected. Among the options provided, the group that would produce a random sample is C) students in every odd-numbered homeroom. Since each odd-numbered homeroom has an equal chance of being selected, the resulting sample would be random and representative of the entire student population. Therefore, the correct answer is C) students in every odd-numbered homeroom.