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irga5000 [103]
3 years ago
6

A basketball team has home games on Thursday and Saturday. The two games together earn $4807.50 for the team. Thursday's game ge

nerates $152.50 less than Saturday's game. How much money was taken in an each game? How much money did Thursday's game generate?
Mathematics
1 answer:
Ket [755]3 years ago
3 0

Answer:

Thursday's game generated $2327.5

Saturday's game generated $2480

explanation:

Let Saturday's games generates x

Then Thursday's game will generate [ x - $152.50 ]

Saturday's + Thursday's = total earnings

  x    +    x - $152.50   =     $4807.50

  2x   =    $4807.50 +  $152.50

  2x   =   $4960

   x    =  $(4960)/2

   x    =   $2480

According to statement, Saturday's games earned $2480

Thursdays game earned = $2480 - $152.50 = $2327.5

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0.997 = 99.7% probability that the resulting sample proportion to be between 0.066 and 0.294

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

18% of residents in Minneapolis, Minnesota, are African Americans. Suppose a local court will randomly sample 100 state residents and will then observe the proportion in the sample who are African American.

This means that p = 0.18, n = 100

So, by the Central Limit Theorem:

\mu = 0.18, s = \sqrt{\frac{0.18*0.82}{100}} = 0.0384

How likely is the resulting sample proportion to be between 0.066 and 0.294?

This is the pvalue of Z when X = 0.294 subtracted by the pvalue of Z when X = 0.066. So

X = 0.294

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

By the Central Limit Theorem

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

Z = \frac{0.294 - 0.18}{0.0384}

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Z = 2.97 has a pvalue of 0.9985

X = 0.066

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

Z = \frac{0.066 - 0.18}{0.0384}

Z = -2.97

Z = -2.97 has a pvalue of 0.0015

0.9985 - 0.0015 = 0.997

0.997 = 99.7% probability that the resulting sample proportion to be between 0.066 and 0.294

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