I have no idea how to them. If someone could do all of them with the work that w

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I have no idea how to them. If someone could do all of them with the work that would be great. Thanks

5. A bank's loan officer 200 and a standard deviation of 50. I that is between 225 and 250. rates applicants for credit. The ratings are normally distributed with a mean of If an applicant is randomly selected, find the probability of a rating (3 pts) 6. Assume that the weights of quarters are normally distributed with a mean of 5.67g and a deviation of 0.070g. A vending machine will only accept coins weighing between deviation of 0,070g. A vending machine will only accept coins weighing between 5.48g and 5.82g What percentage of legal quarters will be rejected? with a mean of 5.67g and a standard (4 pts) 7. I normally distributed with a mean of 1050 kWh anda a) Find Ps , b) If 50 different homes are randomly selected, find the probability that the consumption level for September is greater than 1075 kWh. nomally distrie whe aseetember energy consumption levels for n one region, the September energy consumption levels for single-family homes are found to be single-tamily homes are found to be standard deviation of 218 kWh which is the consumption level separating the bottom 95% from the top 5% t homes are randomly selected, find the probability that the sample mean energy (5 pts) inches, and a standard deviation of 10 inches. What is the probability that the mean annual snowfall duri 25 randomly picked years will exceed 111 inches? The amount of snowfall falling in a certain mountain range is normally distributed with a mean of 109 (3 pts

Explanation / Answer

5.

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    225      
x2 = upper bound =    250      
u = mean =    200      
          
s = standard deviation =    50      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    0.5      
z2 = upper z score = (x2 - u) / s =    1      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.691462461      
P(z < z2) =    0.841344746      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.149882285   [ANSWER]  

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