Louise Johnson
Junior Member
- Joined
- Jan 21, 2007
- Messages
- 103
Question:
Scores on a year-end final Math 10 exam are normally distributed with a mean of 62.5. If the probability of a score being greater than 65.0 is 22%, determine the standard deviation of this distribution, correct to one decimal place.
my answer:
\(\displaystyle \L\\\begin{array}{l}
p\left( {x > 65.0} \right) = 0.22 \\
p\left( {x \le 65.0 = 0.78} \right) \\
p\left( {\frac{{x - 62.5}}{\sigma } \le \frac{{65 - 62.5}}{\sigma }} \right) = .78 \\
p\left( {Z \le \frac{{65 - 62.5}}{\sigma }} \right) = .78invNorm \\
\frac{{2.5}}{\sigma } = .772 \\
\sigma = 3.2 \\
\end{array}\)
Sorry about my answer not being spread out more and easier to read. Lets just say they have come along way. I am dying to know where I am at with this question as I have toiled over it for some time now waiting until I had more to get them posted.
Thank you for your help
Louise
Scores on a year-end final Math 10 exam are normally distributed with a mean of 62.5. If the probability of a score being greater than 65.0 is 22%, determine the standard deviation of this distribution, correct to one decimal place.
my answer:
\(\displaystyle \L\\\begin{array}{l}
p\left( {x > 65.0} \right) = 0.22 \\
p\left( {x \le 65.0 = 0.78} \right) \\
p\left( {\frac{{x - 62.5}}{\sigma } \le \frac{{65 - 62.5}}{\sigma }} \right) = .78 \\
p\left( {Z \le \frac{{65 - 62.5}}{\sigma }} \right) = .78invNorm \\
\frac{{2.5}}{\sigma } = .772 \\
\sigma = 3.2 \\
\end{array}\)
Sorry about my answer not being spread out more and easier to read. Lets just say they have come along way. I am dying to know where I am at with this question as I have toiled over it for some time now waiting until I had more to get them posted.
Thank you for your help
Louise