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Romashka-Z-Leto [24]
2 years ago
7

The amount of money spent weekly on cleaning, maintenance, and repairs at a large restaurant was observed over a long period of

time to be approximately normally distributed, with mean μ = $597 and standard deviation σ = $39.
(a) If $646 is budgeted for next week, what is the probability that the actual costs will exceed the budgeted amount?
Mathematics
1 answer:
Kazeer [188]2 years ago
7 0
Z = ( X - Mean ) / SD
z = ( 646 - 597 ) / 39
z = 49/39 = 1.25
With the z-score table:
for z = 1.25, p = 0.8944 ( probability that the cost will be lower than $646 )
1 - 0.8944 = 0.1056
Answer: <span>The probabilit</span>y is 0.1056
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4 0
2 years ago
A random sample of 23 items is drawn from a population whose standard deviation is unknown. The sample mean isx⎯ ⎯ x¯ = 840 and
gayaneshka [121]

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(832.156, \ 847.844)

Step-by-step explanation:

Given data :

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a). 98% confidence interval

$\overline x \pm t_{(n-1, \alpha /2)}. \frac{s}{\sqrt{n}}$

$E= t_{( n-1, \alpha/2 )} \frac{s}{\sqrt n}}

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