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Rudiy27
2 years ago
5

The meat department of a supermarket sells ground beef in approximate 1 lb packages, but there is some variability. A random sam

ple of 65 packages yielded a mean of 1.05 lbs and a standard deviation of .23 lbs. What is the 90% Confidence Interval for this problem?
Mathematics
1 answer:
grin007 [14]2 years ago
6 0

Answer:

1.05 ± 0.05 lbs

Step-by-step explanation:

Hi!

We can calculate this interval with the z-score of the 90% which is (by convention) 1.645

The interval is calculated as follows:

x_m \pm 1.654(\frac{\sigma}{\sqrt n})

where x_m is the mean, σ the standar deviation and n is the number of samples:

replacing these values we get:

1.05 \pm (1.645 \times \frac{0.23}{\sqrt 65})\\1.05 \pm 0.046

*rounded to the first decimal*

1.05 ± 0.05

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Answer:

4 Sliders 2 chicken wings

Step-by-step explanation:

4x300=1200 2x80=160

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1 year ago
The XO Group Inc. conducted a survey of 13,000 brides and grooms married in the United States and found that the average cost of
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Answer:

a) 0.0392

b) 0.4688

c) At least $39,070 to be among the 5% most expensive.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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:

\mu = 29858, \sigma = 5600

a. What is the probability that a wedding costs less than $20,000 (to 4 decimals)?

This is the pvalue of Z when X = 20000. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{20000 - 29858}{5600}

Z = -1.76

Z = -1.76 has a pvalue of 0.0392.

So this probability is 0.0392.

b. What is the probability that a wedding costs between $20,000 and $30,000 (to 4 decimals)?

This is the pvalue of Z when X = 30000 subtracted by the pvalue of Z when X = 20000.

X = 30000

Z = \frac{X - \mu}{\sigma}

Z = \frac{30000 - 29858}{5600}

Z = 0.02

Z = 0.02 has a pvalue of 0.5080.

X = 20000

Z = \frac{X - \mu}{\sigma}

Z = \frac{20000 - 29858}{5600}

Z = -1.76

Z = -1.76 has a pvalue of 0.0392.

So this probability is 0.5080 - 0.0392 = 0.4688

c. For a wedding to be among the 5% most expensive, how much would it have to cost (to the nearest whole number)?

This is the value of X when Z has a pvalue of 0.95. So this is X when Z = 1.645.

Z = \frac{X - \mu}{\sigma}

1.645 = \frac{X - 29858}{5600}

X - 29858 = 5600*1.645

X = 39070

The wedding would have to cost at least $39,070 to be among the 5% most expensive.

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