A candy bar takes an average of 3.4 hours to move through an assembly line. If the standard deviation is 0.5 hours, what is the probability that the candy bar will take between 3 and 4 hours to move through the line?. A:.2119 B:.2295 C:.6731 D:.3270 E:.8849. 2. the average noise level in a diner is 30 decibels with a standard deviation of 6 decibels. 99% of the time, the noise level is below what value?. A: 16.04 B: 30.00 C: 36.00 D: 43.96 E: 48.00.

Question

Mathematics

Esteban Berry

10 October, 09:21

A candy bar takes an average of 3.4 hours to move through an assembly line. If the standard deviation is 0.5 hours, what is the probability that the candy bar will take between 3 and 4 hours to move through the line?. A:.2119 B:.2295 C:.6731 D:.3270 E:.8849. 2. the average noise level in a diner is 30 decibels with a standard deviation of 6 decibels. 99% of the time, the noise level is below what value?. A: 16.04 B: 30.00 C: 36.00 D: 43.96 E: 48.00.

+5

Answers (1)

Know the Answer?

Josie Rivera · Answer 1

To answer the first problem, we determine first the equivalent z score and get the corresponding probability. The z score at 3 hours is (3-3.4) / 0.5 = 0.8 while that of 4 hrs is (4-3.4) / 0.5 = 1.2. z score of 0.8 is equal to a probability of 0.2119 and that at 1.2 is 0.8849. The difference is 0.673. Hence answer is C.

For problem 2, the z score at 99% is 2.33 hence, the equation becomes 2.33 = (x-30) / 6 where x is the noise level threshold. Answer is D. 43.96 decibels.