find the probability

2011-11-04 7:57 am
7. If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minutes, find the probability that a randomly selected college student will find a parking spot in the library parking lot in less than 3 minutes.
a. 0.0919
b. 0.2255
c. 0.4938
d. 0.7745

8. If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute,m find the probabilty that a randomly selected college student will take between 2 and 4.5 minutes to find a parking spot in the library parking lot.
a. 0.0919
b. 0.2255
c. 0.4938
d. 0.7745

9. If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the point in the distribution in which 75.8% of the college students exceed when trying to find a parking spot in the library parking lot.
a. 2.8 minutes
b. 3.2 minutes
c. 3.4 minutes
d. 4.2 minutes

10. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. A citation catfish should be one of the top 2% in weight. Assuming the weights of catfish are normally distributed, at what weight (in pounds)
should the citation designation be established?
a. 1.56 pounds
b. 4.84 pounds
c. 5.20 pounds
d. 7.36 pound

回答 (1)

2011-11-04 4:40 pm
✔ 最佳答案
The answers is as follow

7 P(X < 3)= P(Z < (3 - 3.5)/1)= P(Z < -0.5)= 0.30858 P(2 < X < 4.5)= P(-1.5 < Z < 1)= 0.7745 (D)9 P(Z > a) = 1 - 0.758a = 0.6999So, the required point is 3.5 + 0.6999 = 4.2 (D)10 P(Z > a) = 0.02a = 2.0537 So,the required point is 3.2 + 2.0537 * 0.8 = 4.84 (B)


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