PLEASE HELP!

1 answer:

5 0

The correlation that is most likely a causation is <span>the positive correlation between the number of sodas sold on a campus and the number of cans in the campus recycling bins.

While the other scenarios may have variables with some relation to each other, in this scenario, we can see that if a campus sells more sodas, they would end up with more cans in the recycling bins; the cans would be caused by the soda sales.</span>

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Lena [83]

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3 years ago

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The average amount parents and children spent per child on back-to-school clothes in Autumn 2010 was $527. Assume the standard d

Ad libitum [116K]

Answer: a. 0.14085

b. 3.826 x $10^{-3}$

c. 0.5437

d. 0.0811

Step-by-step explanation:

Given average amount parents and children spent per child on back-to-school clothes in Autumn 2010 , $\mu$ = $527

Given standard deviation , $\sigma$ = $160

Let X = amount spent on a randomly selected child

Also Z = $\frac{X-\mu}{\sigma}$

a. Probability(X>$700) = P( $\frac{X-\mu}{\sigma}$ > $\frac{700-527}{160}$ ) = P(Z>1.08125) = 0.14085 {Using Z % table}

b. P(X<100) = P( Z < $\frac{100-527}{160}$ ) = P(Z< -2.66875) = P(Z > 2.66875) = 3.826 x $10^{-3}$

c. P(450<X<700) = P(X<700) - P(X<=450)

P(X<700) = 1 - P(X>=700) = 1 - 0.14085 = 0.8592

P(X<=450) = P(Z<= $\frac{450-527}{160}$ ) = P(Z<= -0.48125) = P(Z<=0.48125) = 0.3155

So final P(450<X<700) = 0.8592 - 0.3155 = 0.5437

d. P(X<=300) = P(Z<= $\frac{300-527}{160}$ ) = P(Z<= -1.4188) = P(Z>=1.4188) = 0.0811

All the above probabilities are calculated using Z % table along with interpolation between two values.

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