The normal distribution — practice questions

Let $M$ be the mass of a bird. We have $M \sim N(120, 8^2)$. We want to find $P(110 < M < 130)$.

First, standardise the boundary values: For $M=110$: $Z_1 = \frac{110 - 120}{8} = \frac{-10}{8} = -1.25$ For $M=130$: $Z_2 = \frac{130 - 120}{8} = \frac{10}{8} = 1.25$

So we need to find $P(-1.25 < Z < 1.25)$. This can be calculated as $P(Z < 1.25) - P(Z < -1.25)$.

From the tables, $\Phi(1.25) = P(Z < 1.25) = 0.8944$.

To find $P(Z < -1.25)$, we use symmetry: $P(Z < -1.25) = P(Z > 1.25)$. $P(Z > 1.25) = 1 - P(Z < 1.25) = 1 - 0.8944 = 0.1056$.

Therefore, $P(-1.25 < Z < 1.25) = 0.8944 - 0.1056 = 0.7888$.

The probability is 0.789 (to 3 s.f.).

Alternatively, by symmetry, $P(-1.25 < Z < 1.25) = 2 \times P(0 < Z < 1.25) = 2 \times (\Phi(1.25) - 0.5) = 2 \times (0.8944 - 0.5) = 2 \times 0.3944 = 0.7888$.

Let $X$ be the test score. $X \sim N(\mu, \sigma^2)$.

We are given two pieces of information:

1. $P(X > 75) = 0.15$
2. $P(X < 40) = 0.10$

Standardise both:

1. $P\left(Z > \frac{75 - \mu}{\sigma}\right) = 0.15$. Let $z_1 = \frac{75 - \mu}{\sigma}$. This means $P(Z < z_1) = 1 - 0.15 = 0.85$. From the inverse normal tables (or looking up 0.85 in the probability body), we find $z_1 \approx 1.036$. So, $\frac{75 - \mu}{\sigma} = 1.036$ (Equation A).

2. $P\left(Z < \frac{40 - \mu}{\sigma}\right) = 0.10$. Let $z_2 = \frac{40 - \mu}{\sigma}$. Since the probability is < 0.5, $z_2$ is negative. We find the value for $P(Z < z') = 0.90$, which is $z' \approx 1.282$. So, $z_2 = -1.282$. Thus, $\frac{40 - \mu}{\sigma} = -1.282$ (Equation B).

We now have a pair of simultaneous equations: (A) $75 - \mu = 1.036\sigma$ (B) $40 - \mu = -1.282\sigma$

Subtract (B) from (A): $(75 - \mu) - (40 - \mu) = 1.036\sigma - (-1.282\sigma)$ $35 = 2.318\sigma$ $\sigma = \frac{35}{2.318} = 15.099... \approx 15.1$

Substitute $\sigma$ back into (A): $75 - \mu = 1.036 \times 15.099...$ $75 - \mu = 15.642...$ $\mu = 75 - 15.642... = 59.357... \approx 59.4$

The mean score is 59.4 and the standard deviation is 15.1 (to 3 s.f.).

Let $X$ be the number of heads. $X \sim B(150, 0.4)$. We want to find $P(55 \le X \le 65)$.

First, check conditions for normal approximation: $n=150, p=0.4$. $np = 150 \times 0.4 = 60$. Since $60 > 5$, this is fine. $n(1-p) = 150 \times 0.6 = 90$. Since $90 > 5$, this is also fine. The approximation is appropriate.

Define the normal approximation $Y \sim N(\mu, \sigma^2)$. Mean: $\mu = np = 60$. Variance: $\sigma^2 = np(1-p) = 150 \times 0.4 \times 0.6 = 36$. So, $Y \sim N(60, 36)$ or $Y \sim N(60, 6^2)$.

Apply continuity correction for $P(55 \le X \le 65)$: The range includes 55 and 65. We need to find the area from the lower boundary of 55 to the upper boundary of 65. This becomes $P(54.5 < Y < 65.5)$.

Standardise the boundary values: For $Y=54.5$: $Z_1 = \frac{54.5 - 60}{6} = \frac{-5.5}{6} = -0.9167...$ For $Y=65.5$: $Z_2 = \frac{65.5 - 60}{6} = \frac{5.5}{6} = 0.9167...$

We need to find $P(-0.9167 < Z < 0.9167)$. This is $P(Z < 0.9167) - P(Z < -0.9167)$. From tables, $\Phi(0.9167)$ can be found by interpolating between $\Phi(0.91)=0.8186$ and $\Phi(0.92)=0.8212$. It is approximately $0.8203$. $P(Z < -0.9167) = 1 - P(Z < 0.9167) = 1 - 0.8203 = 0.1797$.

So, the probability is $0.8203 - 0.1797 = 0.6406$. The probability is 0.641 (to 3 s.f.).