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riadik2000 [5.3K]
3 years ago
14

An auction website charges $1 for a bid. The bidding starts at 1¢ and goes up 1¢ at a time. A television that is worth $2000 is

won, on average, with a bid of $160. You make one bid at random. Find the expected value of the outcome of the bid. (Write as an exact decimal, with a negative sign, if necessary.)
Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
3 0
This is fairly complex, but looks like fun. 
So first lets count how many bids are made till the bidding goes up to 160.
That is 100*160=16000
How much will be your random bid. Each bid has equal chance so it is the avarge of all bids. (1cent + 160dollars)/2 = 80dollars (rounded)
What is your chance of winning: 1 bid wins out of 16000 so 1/16000.
When you win your profit is 2000-80dollars=1920
1920/16000= 0.12 dollars 
You entering the bidding costs 1 dollar and your average win is 0.12
This the outcome is 0.12-1=-0.88 dollars
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(Based on Q1 ~ Q3) According to the Bureau of the Census, 18.1% of the U.S. population lives in the Northeast, 21.9% inn the Mid
vekshin1

Answer:

We can therefore conclude that the geographical distribution of hotline callers could be the same as the U.S population distribution.

Step-by-step explanation:

The null Hypothesis: Geographical distribution of hotline callers could be the same as the U.S. population distribution

Alternative hypothesis: Geographical distribution of hotline callers could not be the same as the U.S. population distribution

The populations considered are the Midwest, South, Northeast, and west.

The number of categories, k = 4

Number of recent calls = 200

Let the number of estimated parameters that must be estimated, m = 0

The degree of freedom is given by the formula:

df = k - 1-m

df = 4 -1 - 0 = 3

Let the significance level be, α = 5% = 0.05

For  α = 0.05, and df = 3,

from the chi square distribution table, the critical value = 7.815

<u>Observed and expected frequencies of calls for each of the region:</u>

<u>Northeast</u>

Observed frequency = 39

It contains 18.1% of the US Population

The probability = 0.181

Expected frequency of call = 0.181 * 200 = 36.2

<u>Midwest</u>

Observed frequency = 55

It contains 21.9% of the US Population

The probability = 0.219

Expected frequency of call = 0.219 * 200 =43.8

<u>South</u>

Observed frequency = 60

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Expected frequency of call = 0.367 * 200 = 73.4

<u>West</u>

Observed frequency = 46

It contains 23.3% of the US Population

The probability = 0.233

Expected frequency of call = 0.233 * 200 = 46

x^{2} = \sum \frac{(O_{i} - E_{i})  ^{2} }{E_{i} } ,   i = 1, 2,.........k

Where O_{i} = observed frequency

E_{i} = Expected frequency

Calculate the test statistic value, x²

x^{2} = \frac{(39 - 36.2)^{2} }{36.2} + \frac{(55 - 43.8)^{2} }{43.8} + \frac{(60 - 73.4)^{2} }{73.4} + \frac{(46 - 46.6)^{2} }{46.6}

x^{2} = 5.535

Since the test statistic value, x²= 5.535 is less than the critical value = 7.815, the null hypothesis will not be rejected, i.e. it will be accepted. We can therefore conclude that the geographical distribution of hotline callers could be the same as the U.S population distribution.  

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