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Sliva [168]
3 years ago
8

The United States Department of Agriculture (USDA) found that the proportion of young adults ages 20–39 who regularly skip eatin

g breakfast is 0.238 . Suppose that Lance, a nutritionist, surveys the dietary habits of a random sample of size n = 500 of young adults ages 20–39 in the United States.
Mathematics
1 answer:
FromTheMoon [43]3 years ago
3 0

Answer:

0.3557

Step-by-step explanation:

<em><u>Complete Question:</u></em>

The United States Department of Agriculture (USDA) found that the proportion of young adults ages 20–39 who regularly skip eating breakfast is 0.238 . Suppose that Lance, a nutritionist, surveys the dietary habits of a random sample of size n = 500 of young adults ages 20–39 in the United States.

Apply the central limit theorem for the binomial distribution to find the probability that the number of individuals in Lance's sample who regularly skip breakfast is greater than 122.

<u>Solution:</u>

This is a case where we will use Normal Approximation to Binomial Distribution.

Let X be random variable. So we know:

X ~ Bin (n,p)

Then, Normal Approximation would be:

X ~ Normal Approx (np, npq)

Given in the problem, sample n = 500 and probability of skipping, p = 0.238, so:

X ~ (500, 0.238)

<u>Note:</u> q = 1 - p = 1 - 0.238 = 0.762

Thus,

X ~ Normal Approx (119, 90.678)

Now,

We need P(X > 122).

We convert to Z by using formula:

z=\frac{x-\mu}{\sigma}

Where \mu is the mean, or "np" and \sigma is the standard deviation, which is \sqrt{npq}

THus, we have:

P(X>122)\\=P(\frac{X-\mu}{\sigma}>\frac{122-119}{\sqrt{90.678}})\\=P(Z>0.37)

Which can be said as:

1 - P(Z<0.37)

Using Z-Table (normal table), we have:

1-0.6443=0.3557

THis is our answer.

<u></u>

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Answer:

4.97-2.10\frac{0.0461}{\sqrt{19}}=4.95    

4.97+2.10\frac{0.0461}{\sqrt{19}}=4.99    

So on this case the 95% confidence interval would be given by (4.95;4.99)  

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X =4.97 represent the sample mean

\mu population mean (variable of interest)

s=0.0461 represent the sample standard deviation

n=19 represent the sample size  

Confidence interval

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=19-1=18

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,18)".And we see that t_{\alpha/2}=2.10

Now we have everything in order to replace into formula (1):

4.97-2.10\frac{0.0461}{\sqrt{19}}=4.95    

4.97+2.10\frac{0.0461}{\sqrt{19}}=4.99    

So on this case the 95% confidence interval would be given by (4.95;4.99)    

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The Candela brothers own two pizza restaurants, one on Park Street and one on Bridge Road.
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The mean, median and mode are measures of central tendency, that is they tend to indicate the location middle of the data

Required values;

(a) The performance for the week for Park Street

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  • The sales for the week is better than <u>72.91%</u> of all sales

The performance for the week for Bridge Road

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The median is $<u>3,600</u>

The standard deviation is $<u>3250</u>

The Interquartile range is $<u>6075</u>

Reason:

The table of values that maybe used to find a solution to the question is given as follows;

\begin{array}{|l|l|l|}\mathbf{Variable} &\mathbf{Park}&\mathbf{Bridge}\\N&36&40\\Mean&6611&5989\\SE \ Mean&597&299\\StDev&3580&1794\\Minimum&800&1800\\Q_1&3600&5225\\Median&6600&6000\\Q_3&9675&7625\\Maximum&14100&8600\end{array}\right]

(a) Park Street revenue = $7,500

Bridge Road's revenue = $7,100

The two stores sold close to but below the 75th percentile

Bridge Road revenue;

The z-score is given as follows;

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From the Z-Table, we have;

The percentile= 0.7291

  • Therefore, the sale for the week for Park Street is better than <u>72.91%</u> of all the sales

Park Street revenue;

The z-score is given as follows;

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From the Z-Table, we have;

The percentile = <u>0.5987</u>

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(b) Given that the operating cost is $3,000, frim which we have;

The subtracted value is subtracted from the mean and median to find the new value

Profit = The revenue - Cost

New mean = 6611 - 3000 = 3611

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The new median = 6600 - 3000 = 3600

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The standard deviation and the interquartile range remain the same, therefore, we have;

  • The standard deviation = <u>$3,580</u>

The interquartile range = 9675 - 3600 = 6075

  • The interquartile range = <u>6075</u>

Learn more here:

brainly.com/question/21133077

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