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BigorU [14]
3 years ago
8

Based on a survey conducted by Greenfield Online, 25–34-year-olds spend the most each week on fast food. The average weekly amou

nt of $44 (based on 115 participants) was reported in a May 2009 USA Today Snapshot. Assuming that weekly fast-food expenditures are normally distributed with a known standard deviation of $14.50, construct a 90% confidence interval for the mean weekly amount that 25–34-year-olds spend each week on fast food.
(Please show work)
Mathematics
1 answer:
Artyom0805 [142]3 years ago
8 0
Okay, so the formula for a confidence interval is 
test statistic +/- (critical value)(standard deviation of statistic
44 +/- (invNorm(area: 0.05, u=0, std:1))(14.5)
44 +/- 23.85
(20.15, 67.85)
hope this helps :)
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Answer:

0.5015 = 50.15% probability that it came from manufacturer A.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Defective

Event B: From manufacturer A.

Probability a unit is defective:

2% of 43%(from manufacturer A)

1.5% of 57%(from manufacturer B). So

P(A) = 0.02*0.43 + 0.015*0.57 = 0.01715

Probability a unit is defective and from manufacturer A:

2% of 43%. So

P(A \cap B) = 0.02*0.43 = 0.0086

What is the probability that it came from manufacturer A?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0086}{0.01715} = 0.5015

0.5015 = 50.15% probability that it came from manufacturer A.

4 0
2 years ago
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3 years ago
I need help with this its confusing me too much
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Step-by-step explanation:

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7 0
3 years ago
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Answer:

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