A company manufactures a wide variety of components for use in domestic appliances.

(a) The company introduces a new type of light fitting for refrigerators. Each day a supervisor selects a sample of 100 of the light fittings for testing.

(i) How should the light fittings be selected? Give a reason for this method of selection.

[2]

The supervisor records the number of faulty light fittings found on each of 150 working days. Her results are shown in the table.

(ii) Use the information in the table to estimate $p$, the probability that a light fitting is faulty.

[1]

(iii) Assuming that the number of faulty fittings found each day follows the binomial distribution $B(100, p)$, find the expected number of days on which 3 faulty fittings are found in a period of 150 working days.

[2]

(b) The company also makes heating elements for electric ovens. A fixed number of randomly chosen heating elements are tested each day and the number found to be faulty is denoted by $X$.

(i) State, in context, two assumptions needed for $X$ to be well modelled by a binomial distribution.

Assume now that $X$ has the distribution $\mathrm{B}(80,0.02)$.

[2]

(ii) Find the probability that, on a randomly chosen day, the number of elements found to be faulty is between 1 and 4 inclusive.

[2]

(iii) Find the probability that, in a randomly chosen 5-day working week, more than 3 elements are found to be faulty on at least 2 days.

[3]

(iv) Find the probability that, in a randomly chosen 5-day working week, no more than 8 faulty elements are found in total.

[2]

a(ii)$p=0.03$; (iii)$34.1$ ;

b(ii) $0.779$ ; (iii) $0.0505$ ; (iv) $0.593$

[Maximum marks: 14 marks]

(a)(i) The light fittings should be selected such that he obtains a random sample, i.e., each light fitting has an equal chance of being selected, independently of each other.

(a)(ii) There are several ways to solve this, the best would be to use the GC and calculate using the frequency functions. The probability here is akin to proportion of light fitting that is faulty out of the 150 days of 100 samples each day. So $p=0.03$.

(a)(iii)

$$

\left(\begin{array}{c}

100 \\

3

\end{array}\right) 0.03^3(1-0.03)^{100-3} \times 150 \approx 34.1

$$

Note that the expected number need not be a whole number.

(b)(i) The event that the heating elements for electric ovens is found to be faulty is independent of another heating elements for electric ovens. The probability that heating elements for electric ovens is found to be faulty is constant thoughout the sample.

(b)(ii)

$\mathrm{P}(1 \leq X \leq 4)$

$=\mathrm{P}(X \leq 4)-\mathrm{P}(X=0)$

$\approx 0.779$

(b)(iii)

$$

\mathrm{P}(X>3)=1-\mathrm{P}(X \leq 3)=0.0768549899

$$

Let $Y$ be the number of days with more than 3 elements found to be faulty, out of 5 days.

$$

\begin{array}{l}

Y \sim \mathrm{B}(5,0.0768549899) \\

\mathrm{P}(Y \geq 2)=1-\mathrm{P}(Y \leq 1) \approx 0.0505 \\

\text { (b)(iv) }

\end{array}

$$

Let $W$ be the number of faulty elements found, out of 400 elements.

$$

\begin{array}{l}

W \sim \mathrm{B}(400,0.02) \\

\mathrm{P}(W \leq 8) \approx 0.593

\end{array}

$$