document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1168117>Question 1168117</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The scores for a standardized test follow a normal distribution with a mean test score (the x with the line over it)=84 and standard deviationσ=6.\r<BR>\n" );
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document.write( "a. Approximately 95% of students score between what two numbers?<BR>\n" );
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document.write( "b. What percent of students score below 87? <BR>\n" );
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document.write( "c. What percent of students score above 95? <BR>\n" );
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document.write( "d. What percent of students score between 70 and 90? <BR>\n" );
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document.write( "e. What percent of students score between 95 and 100? <BR>\n" );
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document.write( "f. What score must a student get higher than to be in the top 10%? <BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #851675 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=851675\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let $X$ be the score on the standardized test. We are given that $X$ follows a normal distribution with a mean $\mu = 84$ and a standard deviation $\sigma = 6$.\r<BR>\n" );
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document.write( "**a. Approximately 95% of students score between what two numbers?**\r<BR>\n" );
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document.write( "For a normal distribution, approximately 95% of the data falls within 2 standard deviations of the mean.<BR>\n" );
document.write( "Lower bound = $\mu - 2\sigma = 84 - 2(6) = 84 - 12 = 72$<BR>\n" );
document.write( "Upper bound = $\mu + 2\sigma = 84 + 2(6) = 84 + 12 = 96$\r<BR>\n" );
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document.write( "Approximately 95% of students score between **(72, 96)**.\r<BR>\n" );
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document.write( "**b. What percent of students score below 87?**\r<BR>\n" );
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document.write( "First, we need to find the z-score for a score of 87:<BR>\n" );
document.write( "$z = \frac{x - \mu}{\sigma} = \frac{87 - 84}{6} = \frac{3}{6} = 0.5$\r<BR>\n" );
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document.write( "Now, we need to find the percentage of students scoring below a z-score of 0.5. Looking up the cumulative probability for $z = 0.5$ in a standard normal distribution table or using a calculator, we find $P(Z < 0.5) \approx 0.6915$.\r<BR>\n" );
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document.write( "So, approximately $0.6915 \times 100\% = \boxed{69.15\%}$ of students score below 87.\r<BR>\n" );
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document.write( "**c. What percent of students score above 95?**\r<BR>\n" );
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document.write( "First, we need to find the z-score for a score of 95:<BR>\n" );
document.write( "$z = \frac{x - \mu}{\sigma} = \frac{95 - 84}{6} = \frac{11}{6} \approx 1.83$\r<BR>\n" );
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document.write( "Now, we need to find the percentage of students scoring above a z-score of 1.83. This is $P(Z > 1.83) = 1 - P(Z \le 1.83)$. Looking up the cumulative probability for $z = 1.83$ in a standard normal distribution table or using a calculator, we find $P(Z \le 1.83) \approx 0.9664$.\r<BR>\n" );
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document.write( "So, $P(Z > 1.83) = 1 - 0.9664 = 0.0336$.<BR>\n" );
document.write( "Approximately $0.0336 \times 100\% = \boxed{3.36\%}$ of students score above 95.\r<BR>\n" );
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document.write( "**d. What percent of students score between 70 and 90?**\r<BR>\n" );
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document.write( "First, we need to find the z-scores for scores of 70 and 90:<BR>\n" );
document.write( "For $x = 70$: $z_1 = \frac{70 - 84}{6} = \frac{-14}{6} \approx -2.33$<BR>\n" );
document.write( "For $x = 90$: $z_2 = \frac{90 - 84}{6} = \frac{6}{6} = 1$\r<BR>\n" );
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document.write( "Now, we need to find the percentage of students scoring between z-scores of -2.33 and 1. This is $P(-2.33 < Z < 1) = P(Z < 1) - P(Z \le -2.33)$. Looking up the cumulative probabilities:<BR>\n" );
document.write( "$P(Z < 1) \approx 0.8413$<BR>\n" );
document.write( "$P(Z \le -2.33) \approx 0.0099$\r<BR>\n" );
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document.write( "So, $P(-2.33 < Z < 1) = 0.8413 - 0.0099 = 0.8314$.<BR>\n" );
document.write( "Approximately $0.8314 \times 100\% = \boxed{83.14\%}$ of students score between 70 and 90.\r<BR>\n" );
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document.write( "**e. What percent of students score between 95 and 100?**\r<BR>\n" );
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document.write( "First, we need the z-score for a score of 95 (which we found in part c, $z_1 \approx 1.83$) and the z-score for a score of 100:<BR>\n" );
document.write( "For $x = 100$: $z_2 = \frac{100 - 84}{6} = \frac{16}{6} \approx 2.67$\r<BR>\n" );
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document.write( "Now, we need to find the percentage of students scoring between z-scores of 1.83 and 2.67. This is $P(1.83 < Z < 2.67) = P(Z < 2.67) - P(Z \le 1.83)$. Looking up the cumulative probabilities:<BR>\n" );
document.write( "$P(Z < 2.67) \approx 0.9962$<BR>\n" );
document.write( "$P(Z \le 1.83) \approx 0.9664$\r<BR>\n" );
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document.write( "So, $P(1.83 < Z < 2.67) = 0.9962 - 0.9664 = 0.0298$.<BR>\n" );
document.write( "Approximately $0.0298 \times 100\% = \boxed{2.98\%}$ of students score between 95 and 100.\r<BR>\n" );
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document.write( "**f. What score must a student get higher than to be in the top 10%?**\r<BR>\n" );
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document.write( "We need to find the score $x$ such that the area to the right of $x$ under the normal curve is 0.10. This means the area to the left of $x$ is $1 - 0.10 = 0.90$. We need to find the z-score corresponding to a cumulative probability of 0.90.\r<BR>\n" );
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document.write( "Looking up the z-value for a cumulative probability of 0.90 in a standard normal distribution table or using a calculator, we find $z \approx 1.28$.\r<BR>\n" );
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document.write( "Now, we convert this z-score back to the original score scale:<BR>\n" );
document.write( "$z = \frac{x - \mu}{\sigma}$<BR>\n" );
document.write( "$1.28 = \frac{x - 84}{6}$<BR>\n" );
document.write( "$1.28 \times 6 = x - 84$<BR>\n" );
document.write( "$7.68 = x - 84$<BR>\n" );
document.write( "$x = 84 + 7.68 = 91.68$\r<BR>\n" );
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document.write( "Rounding to the nearest whole number, a student must score higher than $\boxed{92}$ to be in the top 10%.\r<BR>\n" );
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document.write( "Final Answer: The final answer is:<BR>\n" );
document.write( "a. (72, 96)<BR>\n" );
document.write( "b. 69.15%<BR>\n" );
document.write( "c. 3.36%<BR>\n" );
document.write( "d. 83.14%<BR>\n" );
document.write( "e. 2.98%<BR>\n" );
document.write( "f. 92 </SPAN>\n" );
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