Jennys lawn is 1/3 of an acre. after mowing 1/2 of the lawn, how much of an acre has she mowed
1 answer:
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Answer:
a. X~N(2,885, 651)
b. 0.086291
c. 0.00058
d. 3213.10 calories
Step-by-step explanation:
a. -A normal distribution is expressed in the form X~N(mean, standard deviation).
-Let X a random variable denoting the number of calories consumed.
-X is a is a normally distributed random variable with mean 2885 and standard deviation 651.
-This distribution is expressed as X~N(2,885, 651)
b. The probability that less than 2000 calories are consumed is calculated using the formula:
#substitute the given values in the formula to solve for P:
Hence, the probability of consuming less than 2000 calories is 0.08691
c. The proportion of customers consuming more than 5000 calories is calculated as:
Hence, the proportion of customers consuming over 5000 calories is 0.00058
d. The least amount of calories to get the award is calculated as:
1% is equivalent to a z value of 0.50399.
-We equate this to the formula to solve for the mean consumption:
Hence, the least amount of calories consumed to qualify for the award is 3213.10 calories.
