3 years ago

PLEASE help with these 2 questions thank you!

Mathematics

1 answer:

Sergio039 [100]3 years ago

6 0

Download the app photo math help a whole lot

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"0.0373" seems to be the appropriate solution.

Step-by-step explanation:

The given values are:

n₁ = 43

n₂ = 35

$\bar{x_1}=217800$

$\bar{x_2}=204700$

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$\sigma_2=33800$

Now,

The test statistic will be:

⇒ $Z=\frac{\bar{x_1}-\bar{x_2}}\sqrt{\frac{\sigma_1^2}{n_1} +\frac{\sigma_2^2}{n_2} }$

On substituting the given values in the above formula, we get

⇒ $=\frac{217800-204400}{\sqrt{\frac{(30300)^2}{43} +\frac{(33800)^2}{35} } }$

⇒ $=\frac{217800-204400}{\sqrt{\frac{918090000}{43} +\frac{1142440000}{35} } }$

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= $P(Z>1.7828)$

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