Answer:
a) 0.4121
b) $588
Step-by-step explanation:
Mean μ = $633
Standard deviation σ = $45.
Required:
a. If $646 is budgeted for next week, what is the probability that the actual costs will exceed the budgeted amount?
We solve using z score formula
= z = (x-μ)/σ, where
x is the raw score
μ is the population mean
σ is the population standard deviation.
For x = $646
z = 646 - 633/45
z = 0.22222
Probability value from Z-Table:
P(x<646) = 0.58793
P(x>646) = 1 - P(x<646) = 0.41207
≈ 0.4121
b. How much should be budgeted for weekly repairs, cleaning, and maintenance so that the probability that the budgeted amount will be exceeded in a given week is only 0.16? (Round your answer to the nearest dollar.)
Converting 0.16 to percentage = 0.16 × 100% = 16%
The z score of 16%
= -0.994
We are to find x
Using z score formula
z = (x-μ)/σ
-0.994 = x - 633/45
Cross Multiply
-0.994 × 45 = x - 633
-44.73 = x - 633
x = -44.73 + 633
x = $588.27
Approximately to the nearest dollar, the amount should be budgeted for weekly repairs, cleaning, and maintenance so that the probability that the budgeted amount will be exceeded in a given week is only 0.16
is $588
Answer:
.
Step-by-step explanation:
A parallelogram has two pairs of parallel lines so if AD // BC and that AD//BC, Then it is indeed a parallelogram
Answer:
0.0668 = 6.68% of ball bearings has a diameter more than 5.03 mm
Step-by-step explanation:
When the distribution is normal, we use 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.
Manufacturing Ball bearings are manufactured with a mean diameter of 5 millimeters (mm).
This means that
With a standard deviation of 0.02 mm.
This means that
(a) What proportion of ball bearings has a diameter more than 5.03 mm
This is 1 subtracted by the pvalue of Z when X = 5.03. So
has a pvalue of 0.9332
1 - 0.9332 = 0.0668
0.0668 = 6.68% of ball bearings has a diameter more than 5.03 mm
Answer and Step-by-step explanation:
Divide both numbers by 5kg to know how much R is for 1 kg.
= 3.1.
It is 3.1 for 1 kg.
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