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stich3 [128]
3 years ago
13

A teacher decided to bring a jar of 530 pieces of small candy to his 100-student classroom so students could practice estimation

. The students were told that whoever had the closest guess would win the candy. Suppose we took a random sample of one third of the students and calculated the sample mean of their guesses. The distribution of individual guesses had a mean of 400 pieces of candy and a standard deviation of 3,000 pieces of candy (the students had a lot of trouble guessing the count).
Mathematics
1 answer:
nordsb [41]3 years ago
8 0

Answer:

Yes, because the population distribution is normally distributed, and we have a random sample.

Step-by-step explanation:

If the population is skewed, then the sample mean won't be normal for when N is small.

If the population is normal, then the distribution of sample mean looks normal even if N = 2.

If the population is skewed, then the distribution of sample mean looks more and more normal when N gets larger.

Since we have use 33% of the student, then N is large enough to conclude that the student are bad at guessing

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Answer:

2.7

Step-by-step explanation:

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For each expression below, select the letter that corresponds to the equivalent expression given above.
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Part 1)
(x²+15x+65)+(2x-5)*(3x+8)
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Part 2)
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Part 3) 
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Part 4) 
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3 0
4 years ago
How do you find the normal approximation without using the continuity correction?
Dennis_Churaev [7]
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Example: If X is binomially distributed with n=100 and p=0.1, then X has mean np=10 and variance np(1-p)=9. So you can approximate a probability in terms of X with a probability in terms of Y:

\mathbb P(a\le X\le b)\approx\mathbb P(a\le Y\le b)=\mathbb P\left(\dfrac{a-10}3\le\dfrac{Y-10}3\le\dfrac{b-10}3\right)=\mathbb P(a^*\le Z\le b^*)

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3 years ago
You play a game that involves spinning a wheel. Each section of the wheel shown has the same area. Use a sample space to determi
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7 0
3 years ago
Read 2 more answers
Determine the astm grain size number if 33 grains per square inch are measured at a magnification of 270×
strojnjashka [21]
8.9

The equation for the grain size is expressed as the equality:
Nm(M/100)^2 = 2^(n-1)
where
Nm = number of grains per square inch at magnification M.
M = Magnification
n = ASTM grain size number

Let's solve for n, then substitute the known values and calculate.
Nm(M/100)^2 = 2^(n-1)

log(Nm(M/100)^2) = log(2^(n-1))
log(Nm) + 2*log(M/100) = (n-1) * log(2)
(log(Nm) + 2*log(M/100))/log(2) = n-1
(log(Nm) + 2*log(M/100))/log(2) + 1 = n

(log(33) + 2*log(270/100))/log(2) + 1 = n
(1.51851394 + 2*0.431363764)/0.301029996 + 1 = n
(1.51851394 + 0.862727528)/0.301029996 + 1 = n
2.381241468/0.301029996 + 1 = n
7.910312934 + 1 = n
8.910312934 = n

So the ASTM grain size number is 8.9

If you want to calculate the number of grains per square inch, you'd use the
same formula with M equal to 1. So:
Nm(M/100)^2 = 2^(n-1)
Nm(1/100)^2 = 2^(8.9-1)
Nm(1/10000) = 2^7.9
Nm(1/10000) = 238.8564458
Nm = 2388564.458

Or about 2,400,000 grains per square inch.
7 0
3 years ago
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