Question

Packets of milk powder produced by a machine were found to have a normal distribution with a mean mass of 650g and a standard deviation of 10g.
(a) Find the probability that a packet selected at random will have a mass between 620g and 655g.
(b) If 500 packets are selected at random, how many of them will have a mass of more than 660g?
(c) It is found that 10% packets of milk powder will have a mass of less than k grams. Calculate k.

Answer 1

Answer on Question #83464 – Math – Statistics and Probability

Question

Packets of milk powder produced by a machine were found to have a normal distribution with a mean mass of 650g and a standard deviation of 10g.

(a) Find the probability that a packet selected at random will have a mass between 620g and 655g.

(b) If 500 packets are selected at random, how many of them will have a mass of more than 660g?

(c) It is found that 10% packets of milk powder will have a mass of less than k grams. Calculate k.

Solution

(a)

If $X$ is a normally distributed random variable with mean $\mu$ and standard deviation $\sigma$, then the probability that a randomly chosen value of $x$ will be greater than $a$, and less than $b$, is equal to

$$P(a, b \mid \mu, \sigma) = \Phi\left(\frac{b - \mu}{\sigma}\right) - \Phi\left(\frac{a - \mu}{\sigma}\right),$$

where $\Phi(z) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{z} e^{-\frac{t^2}{2}} dt$ is the cumulative distribution function of the standard normal distribution.

Thus,

$$\begin{array}{l} P(620, 655 \mid \mu = 650, \sigma = 10) = \Phi\left(\frac{655 - 650}{10}\right) - \Phi\left(\frac{620 - 650}{10}\right) = 0.691462 - 0.00135 = \\ = 0.690113 \\ \end{array}$$ $$\begin{array}{l} (\text{In Excel } \Phi\left(\frac{655 - 650}{10}\right) - \Phi\left(\frac{620 - 650}{10}\right) = \text{NORMDIST}(655; 650; 10; 1) - \\ \text{NORMDIST}(620; 650; 10; 1)) \\ \end{array}$$

(b)

This value is calculated by the formula

$$N = n * P(x > c / \mu, \sigma) = n(1 - \Phi\left(\frac{c - \mu}{\sigma}\right)),$$

where $n = 500$, $c = 660$.

Thus,

$$N = 500 * (1 - \Phi\left(\frac{660 - 650}{10}\right)) = 500 * (1 - 0.841345) = 500 * 0.158655 = 79$$

(In Excel $\Phi\left(\frac{660 - 650}{10}\right) = \text{NORMDIST}(660; 650; 10; 1)$)

(c)

The value of $k$ is found from the formula $\Phi\left(\frac{k - \mu}{\sigma}\right) = 0.1$:

$$\frac{k - \mu}{\sigma} = \Phi^{-1}(0.1);$$ $$k = \sigma \Phi^{-1}(0.1) + \mu;$$ $$k = 10 * (-1.28155) + 650 = 637$$

(In Excel $\Phi^{-1}(0.1) = \text{NORMSINV}(0.1)$).

Answer:

(a) The probability that a packet selected at random will have a mass between $620\,\mathrm{g}$ and $655\,\mathrm{g}$ is equal to 0.690113.

(b) If 500 packets are selected at random, that 79 of them will have a mass of more than $660\,\mathrm{g}$.

(c) $k = 637\,\mathrm{g}$

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