Solution for #1.a:

Step1: Identify the parts of the problem.

Given:       m = 46.2 𝑚𝑖𝑛𝑢𝑡𝑒𝑠;  𝜎 = 8 𝑚𝑖𝑛𝑢𝑡𝑒𝑠;     𝑋̅ = 43 𝑚𝑖𝑛𝑢𝑡𝑒𝑠;         𝑛 = 50 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠

Find: 𝑃(𝑋̅ < 43)   Step 2: Use the formula to find the z-score.

𝑋̅ −  m

𝒛 = ~~**      s~~       =

√𝑛

43 − 46.2  

8

√50

𝒛 = −𝟐. 𝟖𝟑   Step 3: Use the z-table to look up the z-score you calculated in step 2.

𝒛 = −2.83 has a corresponding area of 0.4977 ****  Step 4: Draw a graph and plot the z-score and its corresponding area. Then, shade the part that you’re looking for: 𝑃(𝑋̅ < 43)

Since we are looking for the probability less than 43 minutes, the shaded part will be on the left part of – 2.83.   Step 5: Subtract your z-score from 0.500.

𝑃(𝑋̅ < 43) = 0.500 − 0.4977

𝑃(𝑋̅ < 43) = 0.0023   Step 6: Convert the decimal in Step 5 to a percentage.

𝑃(𝑋̅ < 43) = 0.23%   \ Therefore, the probability that a randomly selected 50 senior high school students will complete the examination in less than 43 minutes is 0.23%. No, it’s not reasonable since the probability is less than 1.

1.     An electrical company claims that the average life of the bulbs it manufactures is 1 200 hours with a standard deviation of 250 hours. If a random sample of 100 bulbs is chosen, what is the probability that the sample mean will be between 1150 hours and 1 250 hours?   Solution:

Step1: Identify the parts of the problem.

Given:       m = 1200 ℎ𝑜𝑢𝑟𝑠;                𝜎 = 250 ℎ𝑜𝑢𝑟𝑠;           𝑛 = 100 𝑏𝑢𝑙𝑏𝑠

𝑋̅ = 1 150 & 1 250 ℎ𝑜𝑢𝑟𝑠

Unknown: 𝑃(1150 < 𝑋̅ < 1250)   Step 2: Use the formula to find the z-score.   𝒛 =

𝑋̅ − m

~~**      s~~      =

√𝑛

1150 − 1200

250

√100

𝒛 =

1250 − 1200  

250

√100

𝒛 = −𝟐                                                                      𝒛 = 𝟐   Step 3: Use the z-table to look up the z-score you calculated in step 2.

𝒛 = ±2 has a corresponding area of 0.4772 ****  Step 4: Draw a graph and plot the z-score and its corresponding area. Then, shade the part that you’re looking for: 𝑃(1150 < 𝑋̅ < 1250)                 Since we are looking for the probability between 1 150 hours and 1 250 hours, the shaded part will be between –2 and 2.   Step 5: Add the two z-score values.

𝑃(1150 < 𝑋̅ < 1250) = 0.4772 + 0.4772

𝑃(1150 < 𝑋̅ < 1250) = 0.9544   Step 6: Convert the decimal in Step 5 to a percentage.

𝑃(1150 < 𝑋̅ < 1250) = 95.44%   \ Therefore, the probability of randomly selected 100 bulbs to have a sample mean between 1 150 hours and 1 250 hours is 95.44%.

  What’s More   Let’s see how well you understood our discussion. At this point, I want you to solve the following problems. Show your complete solution by following the step-by- step procedure.   1.     The average number of milligrams (mg) of cholesterol in a cup of a certain brand of ice cream is 660 mg, the standard deviation is 35 mg. Assume the variable is normally distributed.

a.      If a cup of ice cream is selected, what is the probability that the cholesterol content will be more than 670 mg?   b.     If a sample of 10 cups of ice cream is selected, what is the probability that the mean of the sample will be larger than 670 mg?