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Naya [18.7K]
3 years ago
8

a random sample of 510 high school students has a normal distribution. The sample mean average ACT exam score was 21 with a 3.2

standard deviation. Construct a 99% confidence interval estimate of the population mean average ACT exam. Find the critical value​
Mathematics
1 answer:
34kurt3 years ago
8 0

Answer:

CI = 21 ± 0.365

Step-by-step explanation:

The confidence interval is:

CI = x ± SE * CV

where x is the sample mean, SE is the standard error, and CV is the critical value (either t score or z score).

Here, x = 21.

The standard error for a sample mean is:

SE = σ / √n

SE = 3.2 / √510

SE = 0.142

The critical value is looked up in a table or found with a calculator.  But first, we must find the alpha level and the critical probability.

α = 1 - 0.99 = 0.01

p* = 1 - (α/2) = 1 - (0.01/2) = 0.995

Using a calculator or a z-score table:

P(x<z) = 0.995

z = 2.576

Therefore:

CI = 21 ± 0.142 × 2.576

CI = 21 ± 0.365

Round as needed.

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Dennis_Churaev [7]

Answer:

M = -2

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Step-by-step explanation:

First we need to find the slope

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m = (-8 -0)/(-1- -5)

   = (-8-0)/(-1+5)

    = -8/4

    = -2

We now want to find the y intercept

y = mx+b is the slope intercept form of the equation

y = -2x+b

Using the point (-5,0)

0 = -2(-5) +b

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Subtract 10 from each side

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The y intercept is -10

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7 0
3 years ago
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Write a word problem whose solution is (-70)=70.
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Step-by-step explanation:

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3 years ago
Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on
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Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

  • If he is given with two kings.
  • If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by =  \frac{^{3}C_2 }{^{47}C_2}   {∵ one king is already there}

              =  \frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}           {∵ ^{n}C_r = \frac{n!}{r! \times (n-r)!} }

              =  \frac{3}{1081}  =  0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by =  \frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}

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Now, probability that he ends up with a full house = \frac{3}{1081} + \frac{6}{1081}

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