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Mashutka [201]
3 years ago
5

Newborn babies in the United States have a mean birth weight of 7.5 pounds and a standard deviation of 1.25 pounds. Assume the d

ata possesses a bell-shaped distribution.
A. What are the upper and lower limits of the interval that contains 95% of all newborns in the United States?

B. Does a newborn with a birth weight of 4.5 pounds fall within an interval which contains 95% of all newborn birth weights. Why or why not?
Mathematics
1 answer:
Aleks [24]3 years ago
6 0

Answer:

A.

Lower limit: 5 pounds

Upper limit: 10 pounds

B.

4.5 is more than two standard deviations from the mean, so it does not fall within an interval which contains 95% of all newborn birth weights.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 7.5

Standard deviation = 1.25

A. What are the upper and lower limits of the interval that contains 95% of all newborns in the United States?

By the Empirical Rule, within 2 standard deviations of the mean.

Lower limit: 7.5 - 2*1.25 = 5 pounds

Upper limit: 7.5 + 2*1.25 = 10 pounds

B. Does a newborn with a birth weight of 4.5 pounds fall within an interval which contains 95% of all newborn birth weights. Why or why not?

4.5 is more than two standard deviations from the mean, so it does not fall within an interval which contains 95% of all newborn birth weights.

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Need pre-cal help. Will mark best answer brainliest
OlgaM077 [116]

so, let's keep in mind that

\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}

so let's make a quick table of those solutions, say A, B, C solutions with x,y,z liters of acid, with an acidity of 0.25, 0.40 and 0.60 respectively.


\bf \begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{liters of }}{amount}\\ \cline{2-4}&\\ A&x&0.25&0.25x\\ B&y&0.40&0.4y\\ C&z&0.60&0.6z\\ \cline{2-4}&\\ mixture&78&0.45&35.1 \end{array} \\\\\\ \begin{cases} x+y+z=78\\ 0.25x+0.4y+0.6z=35.1 \end{cases}


we know she's using "z" liters and those are 3 times as much as "y" liters, so z = 3y.


\bf \begin{cases} x+y+3y=78\\ x+4y=78\\[-0.5em] \hrulefill\\ 0.25x+0.4y+0.6(3y)=35.1\\ 0.25x+0.4y=1.8y=35.1\\ 0.25x+2.2y=35.1 \end{cases}\implies \begin{cases} x+4y=78\\\\ 0.25x+2.2y=35.1 \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ x+4y=78\implies \boxed{x}=78-4y \\\\\\ \stackrel{\textit{using substitution on the 2nd equation}}{0.25\left( \boxed{78-4y} \right)+2.2y=35.1}\implies 19.5-y+2.2y=35.1


\bf 1.2y=15.6\implies y=\cfrac{15.6}{1.2}\implies \blacktriangleright y=13 \blacktriangleleft \\\\\\ x=78-4y\implies x=78-4(13)\implies \blacktriangleright x=26 \blacktriangleleft \\\\\\ z=3y\implies z=3(13)\implies \blacktriangleright z=39 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{25\%}{26}\qquad \stackrel{40\%}{13}\qquad \stackrel{60\%}{39}~\hfill

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