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In the country of United States of Heightlandia, the height measurements of ten-year-old children are approximately normally distributed with a mean of 53.9 inches, and standard deviation of 3 inches.A) What is the probability that a randomly chosen child has a height of less than 47.4 inches?Answer= (Round your answer to 3 decimal places.)B) What is the probability that a randomly chosen child has a height of more than 59.1 inches?Answer= (Round your answer to 3 decimal places.)

User Corstiaan
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Solution:

Given:


\begin{gathered} \mu=53.9 \\ \sigma=3 \\ x=47.4 \end{gathered}

Part A:

Find the Z-score


\begin{gathered} Z=(x-\mu)/(\sigma) \\ Z=(47.4-53.9)/(3) \\ Z=(-6.5)/(3) \\ Z=-2.1667 \end{gathered}

From the Z-scores table, the probability of the height less than 47.4 inches is;


\begin{gathered} P(x<strong>Therefore, the probability that a randomly chosen child has a height of less than 47.4 inches is </strong><strong>0.015</strong><strong> </strong><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p><strong>Part B:</strong></p>[tex]\begin{gathered} \mu=53.9 \\ \sigma=3 \\ x=59.1 \end{gathered}

Find the Z-score


\begin{gathered} Z=(x-\mu)/(\sigma) \\ Z=(59.1-53.9)/(3) \\ Z=(5.2)/(3) \\ Z=1.7333 \end{gathered}

From the Z-scores table, the probability that the height of more than 59.1 inches is;


\begin{gathered} P(x>Z)=0.041521 \\ \\ To\text{ 3 decimal places;} \\ P(x>Z)=0.042 \end{gathered}

Therefore, the probability that a randomly chosen child has a height of more than 59.1 inches is 0.042

User Siddharth Sunchu
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