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4 years ago

7

coffe beans are packaged in 50 kg cartons with an allowable error of 0.7 percent what is the acceptable weight range for a carto

n of coffee beans?

1 answer:

Fofino [41]4 years ago

7 0

Answer:

0.35

Step-by-step explanation:

you mutiply 50 times 0.7%

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5x = 35<br> full answer pls show your work

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5x=35
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A marketing consultant observed 50 consecutive shoppers at a supermarket. One variable of interest was how much each shopper spe

Diano4ka-milaya [45]

Answer:

a. $\\ \mu = USD\;51.34$ ; b. The 95% interval is USD [45.82, 56.86].

Step-by-step explanation:

The mean amount a customer spends at the grocery store

The mean is the sum of all of the data divided by the number of observations for this case, that is, n = 50.

The sum of all of the data is: 5.24+11.97+13.59+15.52+15.69+19.96+20.54+22.59+27.25+27.29+28.25+30.00+31.63+33.01+37.21+40.36+40.90+42.83+45.66+46.67+47.49+49.32+50.12+51.71+52.2 1+53.57+56.16+56.36+56.88+58.65+59.42+60.20+60.48+62.23+62.50+67.95+70.45+73.08+74.15+74.48+74.78+75.75+76.27+76.47+78.92+84.09+84.91+85.06+88.45+88.91 = 2567.17.

The mean $\\ \mu$ is

$\\ \mu = \frac{2567.17}{50}$

$\\ \mu = \frac{2567.17}{50} = 51.34$

Then, <em>the mean amount a customer spends at the grocery store is</em> USD 51.34.

The 95% confidence interval for the mean amount spent

For the 95% confidence interval for the mean, we have the value for the <em>mean = </em>51.34, as we have just found, the value for the<em> standard deviation</em> is already given as USD 20.00. There are 50 observations, so n = 50.

Moreover, for a 95% confidence, the value for the corresponding z-score is z = 1.96 (this is the confidence coefficient). Then, the formula for finding the "limits" values for the confidence interval is as follows:

For the upper limit

$\\ \overline{X} + 1.96*\frac{\sigma}{\sqrt{n}}$

$\\ 51.34 + 1.96*\frac{20}{\sqrt{50}}$

$\\ 51.34 + 5.52$

$\\ 56.86$

For the lower limit

$\\ \overline{X} - 1.96*\frac{\sigma}{\sqrt{n}}$

$\\ 51.34 - 1.96*\frac{20}{\sqrt{50}}$

$\\ 51.34 - 5.52$

$\\ 45.82$

Thus

The 95% confidence is USD [45.82, 56.86].

Mathematically

$\\ P(45.82 \leq \mu \leq 56.86) = 0.95$

There is a probability of 0.95 that the values are between [45.82, 56.86], both inclusive.

We have to remember here that we are dealing with a sample. This sample has a standard deviation, called standard error of the sample, and is $\\ \frac{20}{\sqrt{50}} \approx 2.83$ .

So, the population mean is, with a 95% confidence, between the values [45.82, 56.86], both inclusive.

We can see the graph below that represents the former.

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3 years ago