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yanalaym [24]
3 years ago
14

Scores on an english test are normally distributed with a mean of 31.5 and a standard deviation of 7.3. find the score that sepa

rates the top 59% from the bottom 41%.
Mathematics
1 answer:
kobusy [5.1K]3 years ago
3 0

Let X be the score on an english test which is normally distributed with mean of 31.5 and standard deviation of 7.3

μ = 31.5 and σ =7.3

Here we have to find score that separates the top 59% from the bottom 41%

So basically we have to find here x value such that area above it is 59% and below it is 49%

This is same as finding z score such that probability below z score is 0.49 and above probability is 0.59

P(Z < z) = 0.49

Using excel function to find the z score for probability 0.49 we get

z = NORM.S.INV(0.49)

z = -0.025

It means for z score -0.025 area below it is 41% and above it is 59%

Now we will convert this z score into x value using given mean and standard deviation

x = (z* standard deviation) + mean

x = (-0.025 * 7.3) + 31.5

x = 31.6825 ~ 31.68

The score that separates the top 59% from the bottom 41% is 31.68

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En un videojuego, Marta ha conseguido 36.450 puntos capturando 11 monedas de oro. ¿ cuántos puntos vale cada moneda de oro?
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① 5J+2S =12.20 ② 5J+ 10S =15.80 elimination method
hram777 [196]

Answer:

<h2>S = 9/20</h2><h2>J = 113/50</h2>

Step-by-step explanation:

5J +2S =12.20 ........(1)\\5J+10S=15.80......(2)\\Multiply \:equation \:(1)\:by\:the\:coefficient\:of\:x\:in equation\:(2)\\\\Multiply\:equation(2)\: by\: the\: coefficient\:of \:x\: in \:equation (1)\\\\5J +2S =12.20 ........(1) \times 5\\5J+10S=15.80......(2)\times 5\\\\25J+10S=61......(3)\\25J+50S = 79......(4)\\Subtract\:equation\:(4)\:from\: equation\: (3)\\-40S =-18\\\frac{-40S}{-40}=\frac{-18}{-40}\\  S = 9/20\\

Substitute \:9/20\:for \:x \:in\:equation\:(1)\:or \:equation\:(2)\\5J+2S = 12.20\\5J +2(9/20) =12.20\\5J +9/10=12.20\\5J=12.20-9/10\\5J=113/10\\Cross-Multiply\\50J = 113\\J = 113/50

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3 years ago
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