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3 years ago

At a restaurant you only have $30 to spend on dinner. In addition to the co

Mathematics

1 answer:

mihalych1998 [28]3 years ago

7 0

Answer:

At the restaurant you only have $30 to spend on dinner.

=> 8% sales tax is applied in your bill.

=> And need to leave a 20% tip.

Now, let’s find out how much will be the most expensive item you can order

Let us solve for the 8% tax first

=> 30 dollars * 8%

=> 30 dollars *.08 = 2.4 dollars

Then let’s solve the 20% tip

=> 30 dollars * .20 = 6 dollars

Solve

=> 2.4 + 6.00 = 8.4 dollars

=> 30 dollars – 8.4 dollars = 21.6 dollars

Thus, you can only spend around 22 dollars to be able not to exceed at your 30 dollars budget.

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Use a normal approximation to find the probability of the indicated number of voters. In this case, assume that 104 eligible vot

crimeas [40]

Answer:

The probability that exactly 27 of 104 eligible voters voted is 0.057 = 5.7%.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this case, assume that 104 eligible voters aged 18-24 are randomly selected.

This means that $n = 104$ .

Suppose a previous study showed that among eligible voters aged 18-24, 22% of them voted.

This means that $p = 0.22$

Mean and standard deviation:

$\mu = 104*0.22 = 22.88$

$\sigma = \sqrt{104*0.22*0.78} = 4.2245$

Probability that exactly 27 voted

By continuity continuity, 27 consists of values between 26.5 and 27.5, which means that this probability is the p-value of Z when X = 27.5 subtracted by the p-value of Z when X = 26.5.

X = 27.5

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

$Z = \frac{27.5 - 22.88}{4.2245}$