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The binomial distribution

Practice section 6 of 7

Questions 31–36 in this collection

Practice questions

Review practice · 1 of 6

Translate birthday counts and find a strict minimum sample size

It may be assumed that the dates of birth in a large population are distributed throughout the year so that the probability of a randomly chosen person’s date of birth being in any particular month may be taken as 112\dfrac1{12}.

(i) Find the probability that, of 6 people chosen at random, exactly 2 will have birthdays in January. [3]

(ii) Find the probability that, of 8 people chosen at random, at least one will have a birthday in January. [3]

(iii) NN people are chosen at random. Find the least value of NN such that the probability that at least one will have a birthday in January exceeds 0.9. [3]

(iv) Find the probability that, of 100 people chosen at random, at least 40 will have birthdays in May, June, July or Agusut. [4]

[J84/I/12]

CJC · Review Problems · Question 9

See which counts the words include

(i) Exactly two January birthdays

n = 6, p = 0.08333333333333333. Included counts: 2–2.

Count 0: 0.593; outside eventCount 1: 0.324; outside eventCount 2: 0.0735; in eventCount 3: 0.00891; outside eventCount 4: 0.000608; outside eventCount 5: 0.0000221; outside eventCount 6: 3.35e-7; outside event
Count X ranges from 0 to 6. Use the inspector for individual values.

Highlighted bars form the requested event.

Solution

Review practice · 2 of 6

Derive a cookie success rate and compare two minimum-packet methods

Temasek Bakery produces a large number of “lucky” cookies during the festive season. p%p\% of these cookies contains a winning slip. The cookies are sold by packets. Each packet consists of 8 randomly chosen cookies and the mean number of cookies with a winning slip per packet is 3.2.

(i) Find the probability that a randomly chosen packet contains at most 4 cookies with a winning slip. [2]

(ii) Anand buys 12 packets of cookies. Find the probability that there are more than 10 packets that contain at most 4 cookies with a winning slip. [2]

(iii) Beng decides to buy nn packets of cookies. Find the least value of nn such that the probability that there is at least one packet that contain exactly 4 cookies with a winning slip is more than 0.90. [4]

[2017/TJC/MYE/II/5]

CJC · Review Problems · Question 10

See which counts the words include

(i) At most four slips

n = 8, p = 0.4. Included counts: 0–4.

Count 0: 0.0168; in eventCount 1: 0.0896; in eventCount 2: 0.209; in eventCount 3: 0.279; in eventCount 4: 0.232; in eventCount 5: 0.124; outside eventCount 6: 0.0413; outside eventCount 7: 0.00786; outside eventCount 8: 0.000655; outside event
Count X ranges from 0 to 8. Use the inspector for individual values.

Highlighted bars form the requested event.

Solution

Review practice · 3 of 6

Convert weeks to trials and use a variance restriction

According to the school rules, a student who arrives at school after 0730 hours is considered late.

(i) During the school term, a boy cycles to school on 5 days each week. On any given day, the probability that he arrives after 0730 hours is 0.1. For a period of 4 weeks, calculate the probability that he is late on at least one day but not more than 3 days. [2]

When the boy has cycled to school for the nnth time, the probability that he arrived at school late at least once is greater than 0.99. Calculate the least value of nn. [2]

(ii) A girl travels to school by bus on 5 days a week. Over a long period of time, the variance of the number of days per week on which she is late is 0.8. Given that pp denotes the probability that she arrives late and that p<0.5p<0.5, evaluate pp. [3]

[2007/SRJC/II/10(a)]

CJC · Review Problems · Question 11

See which counts the words include

(i) Between one and three late days

n = 20, p = 0.1. Included counts: 1–3.

Count 0: 0.122; outside eventCount 1: 0.270; in eventCount 2: 0.285; in eventCount 3: 0.190; in eventCount 4: 0.0898; outside eventCount 5: 0.0319; outside eventCount 6: 0.00887; outside eventCount 7: 0.00197; outside eventCount 8: 0.000356; outside eventCount 9: 0.0000527; outside eventCount 10: 0.00000644; outside eventCount 11: 6.51e-7; outside eventCount 12: 5.42e-8; outside eventCount 13: 3.71e-9; outside eventCount 14: 2.06e-10; outside eventCount 15: 9.15e-12; outside eventCount 16: 3.18e-13; outside eventCount 17: 8.31e-15; outside eventCount 18: 1.54e-16; outside eventCount 19: 1.80e-18; outside eventCount 20: 1.00e-20; outside event
Count X ranges from 0 to 20. Use the inspector for individual values.

Highlighted bars form the requested event.

Solution

Review practice · 4 of 6

Read a binomial mode and an expected number of qualifying samples

Samples, each of 12 articles, are taken at random from a large consignment in which 25% of articles are defective.

Find the number of defective articles which is most likely to occur in a single sample, and find the probability of obtaining this number.

If 100 samples of 12 articles are to be examined, calculate the number of samples in which you would expect to find 4 or more defective articles.

CJC · Review Problems · Question 12

See which counts the words include

Most likely count: three defectives

n = 12, p = 0.25. Included counts: 3–3.

Count 0: 0.0317; outside eventCount 1: 0.127; outside eventCount 2: 0.232; outside eventCount 3: 0.258; in eventCount 4: 0.194; outside eventCount 5: 0.103; outside eventCount 6: 0.0401; outside eventCount 7: 0.0115; outside eventCount 8: 0.00239; outside eventCount 9: 0.000354; outside eventCount 10: 0.0000354; outside eventCount 11: 0.00000215; outside eventCount 12: 5.96e-8; outside event
Count X ranges from 0 to 12. Use the inspector for individual values.

Highlighted bars form the requested event.

Solution

Review practice · 5 of 6

Use a noninteger mean, compare modal counts and audit finite-population assumptions

One question in a survey is as follows:

“Do you use Facebook during office hours at least once a day? Yes or No?”

It is known that on average, 7 out of 9 staff members answered “yes” to the above question.

For a random sample of 60 staff members,

(i) find the standard deviation of the number of staff members who answer “yes”, [1]

(ii) find the probability that the number of staff members who answer “yes” exceeds the mean, [3]

(iii) find the most probable number of staff members who answer “yes”, [2]

(iv) state, with a reason, if the above answers are valid if the total number of staff members in the company is small. [1]

[2013/DHS/10(b)(modified)]

CJC · Review Problems · Question 13

See which counts the words include

(ii) More yes responses than the mean

n = 60, p = 0.7777777777777778. Included counts: 47–60.

Count 0: 6.42e-40; outside eventCount 1: 1.35e-37; outside eventCount 2: 1.39e-35; outside eventCount 3: 9.41e-34; outside eventCount 4: 4.69e-32; outside eventCount 5: 1.84e-30; outside eventCount 6: 5.90e-29; outside eventCount 7: 1.59e-27; outside eventCount 8: 3.70e-26; outside eventCount 9: 7.48e-25; outside eventCount 10: 1.33e-23; outside eventCount 11: 2.12e-22; outside eventCount 12: 3.03e-21; outside eventCount 13: 3.92e-20; outside eventCount 14: 4.61e-19; outside eventCount 15: 4.94e-18; outside eventCount 16: 4.87e-17; outside eventCount 17: 4.41e-16; outside eventCount 18: 3.69e-15; outside eventCount 19: 2.85e-14; outside eventCount 20: 2.05e-13; outside eventCount 21: 1.36e-12; outside eventCount 22: 8.47e-12; outside eventCount 23: 4.90e-11; outside eventCount 24: 2.64e-10; outside eventCount 25: 1.33e-9; outside eventCount 26: 6.27e-9; outside eventCount 27: 2.76e-8; outside eventCount 28: 1.14e-7; outside eventCount 29: 4.40e-7; outside eventCount 30: 0.00000159; outside eventCount 31: 0.00000539; outside eventCount 32: 0.0000171; outside eventCount 33: 0.0000508; outside eventCount 34: 0.000141; outside eventCount 35: 0.000367; outside eventCount 36: 0.000893; outside eventCount 37: 0.00203; outside eventCount 38: 0.00429; outside eventCount 39: 0.00848; outside eventCount 40: 0.0156; outside eventCount 41: 0.0266; outside eventCount 42: 0.0421; outside eventCount 43: 0.0617; outside eventCount 44: 0.0834; outside eventCount 45: 0.104; outside eventCount 46: 0.118; outside eventCount 47: 0.124; in eventCount 48: 0.117; in eventCount 49: 0.100; in eventCount 50: 0.0773; in eventCount 51: 0.0530; in eventCount 52: 0.0321; in eventCount 53: 0.0170; in eventCount 54: 0.00770; in eventCount 55: 0.00294; in eventCount 56: 0.000919; in eventCount 57: 0.000226; in eventCount 58: 0.0000408; in eventCount 59: 0.00000485; in eventCount 60: 2.83e-7; in event
Count X ranges from 0 to 60. Use the inspector for individual values.

Highlighted bars form the requested event.

Solution

Review practice · 6 of 6

Find the smallest binomial distribution with two tied modes

The random variable XX is the number of successes in nn independent trials of an experiment in which the probability of a success at any trial is pp.

Denoting P(X=k)P(X=k) by pkp_k, it is given that pkpk−1=(n−k+1)pk(1−p)\dfrac{p_k}{p_{k-1}}=\dfrac{(n-k+1)p}{k(1-p)}, k=1,2,…,nk=1,2,\ldots,n.

A distribution is said to be bimodal if it has two modes.

Find the least value of nn, and the corresponding modes of XX, given that XX is bimodal and that p=1825p=\dfrac{18}{25}. [3]

[2017/ACJC/MYE/I/8]

CJC · Review Problems · Question 14

See which counts the words include

First tied mode: 17

n = 24, p = 0.72. Included counts: 17–17.

Count 0: 5.39e-14; outside eventCount 1: 3.33e-12; outside eventCount 2: 9.84e-11; outside eventCount 3: 1.86e-9; outside eventCount 4: 2.51e-8; outside eventCount 5: 2.58e-7; outside eventCount 6: 0.00000210; outside eventCount 7: 0.0000139; outside eventCount 8: 0.0000758; outside eventCount 9: 0.000347; outside eventCount 10: 0.00134; outside eventCount 11: 0.00438; outside eventCount 12: 0.0122; outside eventCount 13: 0.0289; outside eventCount 14: 0.0584; outside eventCount 15: 0.100; outside eventCount 16: 0.145; outside eventCount 17: 0.175; in eventCount 18: 0.175; outside eventCount 19: 0.142; outside eventCount 20: 0.0915; outside eventCount 21: 0.0448; outside eventCount 22: 0.0157; outside eventCount 23: 0.00352; outside eventCount 24: 0.000377; outside event
Count X ranges from 0 to 24. Use the inspector for individual values.

Highlighted bars form the requested event.

Solution

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