Mathski
Worksheet
Normal Distribution II
IB Mathematics AA · SL & HL · Shadow Worksheet
Practice
Name
Question 1
\(X\sim N(80,25)\). Find the value of \(a\) such that \(P(X
Inverse normal
\(P(Z
Back-transform
\(a=80+1.282\times5\)
Answer
\(a\approx86.4\)
Question 2
\(X\sim N(\mu,\sigma^2)\). Given \(P(X>90)=0.10\) and \(P(X<65)=0.05\), find \(\mu\) and \(\sigma\).
Z-scores
\(\tfrac{90-\mu}{\sigma}=1.282\); \(\tfrac{65-\mu}{\sigma}=-1.645\)
Subtract equations
\(\tfrac{25}{\sigma}=2.927\Rightarrow\sigma\approx8.54\)
Find \(\mu\)
\(\mu=90-1.282\times8.54\approx79.0\)
Answer
\(\mu\approx79.0,\;\sigma\approx8.54\)
Question 3
Exam scores are normally distributed with mean 62 and standard deviation 11. The top 15% get a distinction. Find the minimum score for a distinction.
Top 15% means \(P(X>a)=0.15\Rightarrow P(Z>z)=0.15\Rightarrow z\approx1.036\)
Back-transform
\(a=62+1.036\times11\approx73.4\)
Answer
Minimum score \(\approx74\) (rounding up)
Question 4
The lifetimes of batteries are \(N(120,\sigma^2)\) hours. Given 5% last less than 100 hours, find \(\sigma\) and the probability a battery lasts more than 140 hours.
Find \(\sigma\)
\(\tfrac{100-120}{\sigma}=-1.645\Rightarrow\sigma=\tfrac{20}{1.645}\approx12.2\)
\(P(X>140)\)
\(Z=\tfrac{140-120}{12.2}\approx1.64\); \(P(Z>1.64)\approx0.0505\)
Answer
\(\sigma\approx12.2\); \(P(X>140)\approx0.0505\)
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