1. Let $X$ be a normal random variable with $\mu = 6$ and $\sigma = 5$. Find $P( 10 < X < 20)$.

    $0.8159 - 0.6554 = 0.1605$ $0.758 - 0.5793 = 0.1788$ $0.9998 - 0.9452 = 0.0546$ $0.9974 - 0.7881 = 0.2093$

  2. Consider a sample $X_1,\ldots,X_{ 9}$ of normally distributed random variables with mean $\mu = 2$ and variance $\sigma^2 = 4$. What is the probability that $1.6 \le \bar{X} \le 2.4$?

    $\Phi( 0.13) - \Phi( -0.13) \approx 0.1061$ $\Phi( 0.4) - \Phi( -0.4) \approx 0.3108$ $\Phi( 0.15) - \Phi( -0.15) \approx 0.1192$ $\Phi( 2.4) - \Phi( -2.4) \approx 0.9836$ $\Phi( 0.6) - \Phi( -0.6) \approx 0.4515$

  3. Let $Z$ be a standard normal random variable. Find the value of $x$ such that $P(Z > x) = 0.01$.

    $2.33$ $1.16$ $0.82$ $1.64$

  4. Let $X$ be a normal random variable with $\mu = 6$ and $\sigma = 5$. Find $P(X > 5)$.

    $1 - 0.4602$ $1 - 0.7257$ $1 - 0.4207$ $0.4207$ $0.4602$ $0.7257$

  5. An actual voltage of new battery has the mean value of $3$ and the variance ${{4}\over{3}}$. Consider the the average voltages from 30 new batteries. Find the mean and the variance.

    The mean is ${{1}\over{10}}$ and the variance is $30$ The mean is $3$ and the variance is ${{2}\over{45}}$ The mean is $3$ and the variance is $30$ The mean is ${{1}\over{10}}$ and the variance is ${{1}\over{30}}$ The mean is $3$ and the variance is ${{1}\over{30}}$ The mean is ${{1}\over{10}}$ and the variance is ${{2}\over{45}}$


Department of Mathematics
Last modified: 2026-07-14