Anticipated consumer demand in a restaurant for free-range steaks next month can be modeled by a normal random variable with mean 1,200 pounds and standard deviation 100 pounds.
a. What is the probability that demand will exceed 1,000 pounds?
b. What is the probability that demand will be between 1,100 and 1,300 pounds?
c. The probability is 0.10 that demand will be more than how many pounds?
Answers
Given : modeled by a normal random variable with mean 1,200 pounds and standard deviation 100 pounds.
To find : probability that demand will exceed 1,000 pounds. probability that demand will be between 1,100 and 1,300 pounds? The probability is 0.10 that demand will be more than how many pounds
Solution:
Mean = 1200
Standard Deviation = 100
Z score = ( Value - Mean) / Standard Deviation
Value = 1000
=> Z score = (1000 - 1200)/100 = -2
-2 Z score = 0.0228
probability that demand will exceed 1,000 pounds = 1 - 0.0228 = 0.9772
1,100 and 1,300
=> Z score = - 1 & 1
0.1587 , 0.8413
probability that demand will be between 1,100 and 1,300 pounds = 0.8413 - 0.1587 = 0.6826
The probability is 0.10 => 1 - 0.1 = 0.9
=> z score = 1.281
1.281 = ( value -1200)/100
=> Value = 1328.1
demand will be more 1328.1 pounds
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