Question

Please need some help on this
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

Answer 1

use this to estimate the area under the curve. Remember that 1on teh axis means one standard deviation

Answer 2

and zero is the mean

Answer 3

The first part of the question can be solved by finding the z-scores for 3 hours and 4 hours as follows:
\[z _{(4\ hours)}=\frac{4-3.4}{0.5}=1.2\]
\[z _{(3\ hours)}=\frac{3-3.4}{0.5}=-0.8\]
A standard normal distribution table is then used to find the cumulative probabilities for these two z-scores which are:
P(4 hours) = 0.8849
P(3 hours) = 0.2119
The required probability = 0.8849 - 0.2119 = you can calculate.

For the second part of the question, first use a standard normal distribution table to find the z-score for a cumulative probability of 99%. This z-score is 2.33. Let X be the required value of noise level. Then we have
\[z _{(for\ p= 99\ percent)}=2.33=\frac{X-30}{6}\ ..........(1)\]
By rearranging equation (1) we can find the value of X as follows:
\[X=30+(6\times2.33)= you\ can\ calculate\]