The girls in Lana’s troop set a goal to sell 1,000 boxes of cookies this year. There are 13 girls in the troop. At least how man
2 answers:
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Answer:
a) 94.06% probability that a water specimen collected from the river has an alkalinity level exceeding 45 milligrams per liter.
b) 94.06% probability that a water specimen collected from the river has an alkalinity level below 55 milligrams per liter.
c) 50.98% probability that a water specimen collected from the river has an alkalinity level between 48 and 52 milligrams per liter.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
a. exceeding 45 milligrams per liter.
This probability is 1 subtracted by the pvalue of Z when X = 45. So
has a pvalue of 0.0594.
1 - 0.0594 = 0.9406
94.06% probability that a water specimen collected from the river has an alkalinity level exceeding 45 milligrams per liter.
b. below 55 milligrams per liter.
This probability is the pvalue of Z when X = 55.
has a pvalue of 0.9604.
94.06% probability that a water specimen collected from the river has an alkalinity level below 55 milligrams per liter.
c. between 48 and 52 milligrams per liter.
This is the pvalue of Z when X = 52 subtracted by the pvalue of Z when X = 48. So
X = 52
has a pvalue of 0.7549
X = 48
has a pvalue of 0.2451
0.7549 - 0.2451 = 0.5098
50.98% probability that a water specimen collected from the river has an alkalinity level between 48 and 52 milligrams per liter.