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vfiekz [6]
3 years ago
9

Some sources report that the weights of​ full-term newborn babies in a certain town have a mean of 8 pounds and a standard devia

tion of 0.6 pounds and are normally distributed. a. What is the probability that one newborn baby will have a weight within 0.6 pounds of the meaning dash that ​is, between 7.4 and 8.6 ​pounds, or within one standard deviation of the​ mean
Mathematics
1 answer:
Elodia [21]3 years ago
7 0

Answer:

0.682 is the probability that one newborn baby will have a weight within one standard deviation of the​ mean.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  8 pounds

Standard Deviation, σ = 0.6 pounds

We are given that the distribution of weights of​ full-term newborn babies is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P(weight between 7.4 and 8.6 ​pounds)

P(7.4 \leq x \leq 8.6) = P(\displaystyle\frac{7.4 - 8}{0.6} \leq z \leq \displaystyle\frac{8.6-8}{0.6}) = P(-1 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -1)\\= 0.841 - 0.159 = 0.682 = 68.2\%

P(7.4 \leq x \leq 8.6) = 6.82\%

This could also be found with the empirical formula.

0.682 is the probability that one newborn baby will have a weight within 0.6 pounds of the mean.

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5x+8y=-9\implies 8y=-5x-9\implies y=\cfrac{-5x-9}{8} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{5}{8}} x-9\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}

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