1. Let $X$ be a normal random variable with $\mu = 2$ and $\sigma = 10$. Find $P( 5 < X < 10)$.

    $0.7881 - 0.6179 = 0.1702$ $0.8849 - 0.5793 = 0.3057$ $0.7257 - 0.5398 = 0.1859$ $0.9452 - 0.7257 = 0.2195$

  2. Let $X$ be a normal random variable with $\mu = 2$ and $\sigma = 10$. Find $P(X > 0)$.

    $1 - 0.2743$ $0.4602$ $0.4207$ $1 - 0.4602$ $0.2743$ $1 - 0.4207$

  3. Let $Z$ be a standard normal random variable. Find the value of $x$ such that $P(Z > x) = 0.05$.

    $0.64$ $0.82$ $1.64$ $1.28$

  4. Consider a sample $X_1,\ldots,X_{ 25}$ of normally distributed random variables with mean $\mu = 2$ and variance $\sigma^2 = 1$. What is the probability that $1.8 \le \bar{X} \le 2.2$?

    $\Phi( 0.5) - \Phi( -0.5) \approx 0.3829$ $\Phi( 1.0) - \Phi( -1.0) \approx 0.6827$ $\Phi( 0.2) - \Phi( -0.2) \approx 0.1585$ $\Phi( 0.04) - \Phi( -0.04) \approx 0.0319$ $\Phi( 2.0) - \Phi( -2.0) \approx 0.9545$

  5. An actual voltage of new battery has the mean value of $3$ and the variance ${{1}\over{3}}$. Consider the the average voltages from 30 new batteries. Find the mean and the variance.

    The mean is $3$ and the variance is ${{1}\over{90}}$ The mean is $3$ and the variance is $30$ The mean is $3$ and the variance is ${{1}\over{30}}$ The mean is ${{1}\over{10}}$ and the variance is ${{1}\over{30}}$ The mean is ${{1}\over{10}}$ and the variance is $30$ The mean is ${{1}\over{10}}$ and the variance is ${{1}\over{90}}$

  6. An actual voltage of new battery has the mean value of $9$ and the variance ${{4}\over{3}}$. Consider the average voltage $Y$ of $27$ new batteries. Approximate $P( {{26}\over{3}} < Y < {{28}\over{3}})$.

    $\Phi( {{9}\over{8}}) - \Phi( -{{9}\over{8}}) \approx 0.7394$ $\Phi( {{1}\over{8}}) - \Phi( -{{1}\over{8}}) \approx 0.0995$ $\Phi( {{1}\over{6}}) - \Phi( -{{1}\over{6}}) \approx 0.1324$ $\Phi( {{3}\over{2}}) - \Phi( -{{3}\over{2}}) \approx 0.8664$

  7. The germination time in days of a newly planted seed has the mean value of $2$ and variance $1$. If the germination times $X_1,\ldots,X_n$ are independent, find the mean and the variance for the average germination time $\bar{X}$ of $n = 100$ seeds.

    The mean is $8$ and the variance is ${{1}\over{25}}$ The mean is $2$ and the variance is $1$ The mean is $8$ and the variance is ${{1}\over{100}}$ The mean is $2$ and the variance is ${{1}\over{25}}$ The mean is $8$ and the variance is $1$ The mean is $2$ and the variance is ${{1}\over{100}}$

  8. The germination time in days of a newly planted seed has the mean value of $18$ and variance $36$. If the germination times $X_1,\ldots,X_n$ are independent, estimate the probability that the average germination time $\bar{X}$ of $n = 36$ seeds is between 17.9 and 19.6 days.

    $\Phi( 2.17) - \Phi( -0.67) \approx 0.7324$ $\Phi( 1.3) - \Phi( -0.4) \approx 0.5586$ $\Phi( 1.6) - \Phi( -0.1) \approx 0.485$ $\Phi( 2.67) - \Phi( -0.17) \approx 0.5624$


Department of Mathematics
Last modified: 2026-06-07