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Aleksandr-060686 [28]
3 years ago
7

3. The three salespeople for a local advertising firm are Lola, Ahmed, and Tommy. Lola sold $2030 in ads, Ahmed sold $1540, and

Tommy sold $1800. (a) What fraction of the total sales did each salesperson sell? (b) A $100 bonus was awarded to the three salespeople, which they had to share. It was awarded so that each salesperson received the same fraction of $100 as he or she sold of the total sales. How much did each person receive? Round to the nearest whole cent.
Mathematics
1 answer:
dalvyx [7]3 years ago
6 0
A.) First, you find the total sales. $2030 + $1540 + $1800 = $5370. Then, divide each with the total. The answer would be:

Lola: 203/537
Ahmed: 154/537
Tommy: 60/179

b.) Divide 100 by 3 for equal shares. Each person receives $33.3.
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Answer:

a. If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b. If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

Step-by-step explanation:

This problem should be solved with a binomial distribution sample, but as the size of the sample is large, it can be approximated to a normal distribution.

The parameters for the normal distribution will be

\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/200}= 0.0353

We can calculate the z values for x1=0.47 and x2=0.51:

z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.0353}=-0.85\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.0353}=0.28

We can now calculate the probabilities:

P(0.47

If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b) If the sample size change, the standard deviation of the normal distribution changes:

\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/100}= 0.05

We can calculate the z values for x1=0.47 and x2=0.51:

z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.05}=-0.6\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.05}=0.2

We can now calculate the probabilities:

P(0.47

If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

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