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Worked solution

Estimate a defect probability from a frequency table

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

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.

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

Number faultyNumber of days
04
119
238
341
422
516
66
74
8 or more0

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

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

[2021/II/11(a)]

CJC · Discussion · Question 10

Answer

(i) Random selection with equal chances, to avoid selection bias. (ii) p=0.03p=0.03. (iii) 34.134.1 days.

(i)

Take a random sample, giving each fitting an equal chance of selection, to avoid systematic selection bias. The source also specifies independent selections; independence/common defect probability are modelling assumptions supporting the later binomial calculation. For a finite sample selected without replacement, equal selection chances alone do not imply exact independence.

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(ii)

The total faults recorded are 0(4)+1(19)+2(38)+3(41)+4(22)+5(16)+6(6)+7(4)=4500(4)+1(19)+2(38)+3(41)+4(22)+5(16)+6(6)+7(4)=450. The frequencies total 150 days; the 8-or-more category has frequency zero, so no missing tail value is needed. Mean faults per day =450/150=3. Since np=3 with n=100, the estimate is p=0.03.

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(iii)

Let X∼B(100,0.03)X\sim B(100,0.03) count faulty fittings in a daily sample. Then P(X=3)=(1003)(0.03)3(0.97)97=0.2274741275…P(X=3)=\binom{100}3(0.03)^3(0.97)^{97}=0.2274741275\ldots. The expected number of days is 150 times this probability: 34.12111912…=34.1 days (3 s.f.). An expected frequency need not be an integer.

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Common mistakes

The table counts days, not fittings. The zero-frequency “8 or more” category must be retained.

Solution notes

The source table is transposed in the student presentation for narrow screens; every category and frequency is unchanged. The original two-row table is retained in the native record.

Source sampling wording includes independent selection. The reviewed answer distinguishes that model premise from equal-chance finite sampling without replacement.

Read the hint