1.Calculate the GPA of a student with the following grades: B (11hours), D (4 ho
ID: 3295816 • Letter: 1
Question
1.Calculate the GPA of a student with the following grades: B (11hours), D (4 hours), F (9hours). Note that an A is equivalent to 4.0, a B is equivalent to a 3.0, a C is equivalent to a 2.0, a D is equivalent to a 1.0, and an F is equivalent to a 0. Round your answer to two decimal places.
2.The life of light bulbs is distributed normally. The variance of the lifetime is 225and the mean lifetime of a bulb is 590hours. Find the probability of a bulb lasting for at most 624 hours. Round your answer to four decimal places.
3. A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean life of 75 months with a standard deviation of 5 months. If the claim is true, what is the probability that the mean monitor life would be greater than 74.4 months in a sample of 85 monitors? Round your answer to four decimal places.
Explanation / Answer
1. Add up all the grade points corresponding to the grades B, D and F, that is 4 (3+1+0). The number of class credits taken, that is 24 (11+4+9). Divide the added grade points by the total number of class credits taken. Therefore, the GPA is: 4/24=0.17
2. From information given, mu=590, sigma=sqrt 225=15. The lif eof light bulb is normally distributed, therefore, use the Z score formula to compute the probability. Assume X denote the life time of a bulb in hours.
P(X<=624)=P[Z<=(624-590)/15] [Use formula Z=(X-mu)/sigma, where, X is the raw score, mu is the population mean, and sigma is population standard deviation].
=P[Z<=2.27] (look into area in the z table corresponding to the z score. The z table gives area under standard normal curve to the left of z).
=0.9884
3. From information given, mu=75, n=85, xbar=74.4, sigmaxbar=sigma/sqrt n=5/sqrt 85=0.5423.
Substitute the values in the z score formula and compute the required probability.
P(xbar>74.4)=P[Z>(74.4-75)/0.5423]=P(Z>-1.11) [z table gives area to the left of z, therefore, to find area greater than a particular z score, subtract the area from 1].
=1-0.1335=0.8665
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.