I'm having a little bit of trouble figuring out the formula to solve this problem.
1 answer:
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Mean weight of the bag of pears = u = 8 pounds
Standard deviation = s = 0.5 pounds
We have to find what percentage of bags of pears will weigh more than 8.25 pounds. This can be done using the z score.
We have to convert x = 8.25 to z scores, which will be:
z score = 0.5
From the z table, the probability of z score being greater than 0.5 is 0.3085
Therefore, the probability of a bag to weigh more than 8.25 pounds is 0.3085
Thus 0.3085 or 31% (rounded to nearest integer) of bags of pears will have weight more than 8.25 pounds at the local market.
Standard deviation = s = 0.5 pounds
We have to find what percentage of bags of pears will weigh more than 8.25 pounds. This can be done using the z score.
We have to convert x = 8.25 to z scores, which will be:
z score = 0.5
From the z table, the probability of z score being greater than 0.5 is 0.3085
Therefore, the probability of a bag to weigh more than 8.25 pounds is 0.3085
Thus 0.3085 or 31% (rounded to nearest integer) of bags of pears will have weight more than 8.25 pounds at the local market.
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