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snow_lady [41]
3 years ago
15

A random sample of 15 statistics textbooks has a mean price of $105 with a standard deviation of $30.25. Determine whether a nor

mal distribution or a t-distribution should be used or whether neither of these can be used to construct a confidence interval. Assume the distribution of statistics textbook prices is not normally distributed
Mathematics
1 answer:
Katena32 [7]3 years ago
7 0

Answer:

neither normal distribution nor t-distribution

Step-by-step explanation:

given data

no of sample = 15

mean = $105

standard deviation = $30.25

to find out

whether a normal distribution or a t-distribution

solution

this is neither normal distribution nor t- distribution because here we know t, z is use for normal and difference ( n greater  than 30 and Standard Deviation used z  and the  for n less than 30 ,  Standard Deviation not known use for t distribution  ) so it we can say it is neither normal distribution nor t- distribution

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In the Figure below is shown the graph of this function. We have the following function:

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The y-intercept occurs when x=0, so:

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Therefore, the y-intercept is the given by the point:

\boxed{(0,-2)}

From the figure we have three x-intercepts:

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So, the x-intercepts occur when y=0. Thus, proving this:

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3 years ago
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a = 7 , b = -1 , c = -9

x = \frac{-b \pm  \sqrt{b^2 - 4ac} }{2a}   Plug in the a, b, and c values
x = \frac{- (-1) \pm  \sqrt{(-1)^2 - 4(7)(-9)} }{2(7)}   Cancel out the double negative
x = \frac{1 \pm  \sqrt{(-1)^2 - 4(7)(-9)} }{2(7)}   Square -1
x = \frac{1 \pm  \sqrt{1 - 4(7)(-9)} }{2(7)}   Multiply 7 and -9
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x = \left \{ {{ \frac{1 +  \sqrt{253} }{14} } \atop { \frac{1 -  \sqrt{253} }{14} }} \right.
The approximate square root of 253 is <span>15.905973.
</span>x ≈ \left \{ { \frac{1 + 15.905973}{14} } \atop { \frac{1 - 15.905973}{14} }} \right   Add and subtract
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x ≈ \left \{ {{1.2075} \atop {1.0647}} \right.   Round to the nearest hundredth
x ≈ \left \{ {{1.21} \atop {1.06}} \right.

<span>
</span>
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