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Patrick is practising his skateboarding skills - AQA - A-Level Maths Pure - Question 13 - 2019 - Paper 3

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Patrick is practising his skateboarding skills. On each day, he has 30 attempts at performing a difficult trick. Every time he attempts the trick, there is a probab... show full transcript

Worked Solution & Example Answer:Patrick is practising his skateboarding skills - AQA - A-Level Maths Pure - Question 13 - 2019 - Paper 3

Step 1

Find the mean number of times he falls off in a day.

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Answer

To find the mean of a binomial distribution, we use the formula:

extMean=nimesp ext{Mean} = n imes p

where:

  • n=30n = 30 (number of attempts)
  • p=0.2p = 0.2 (probability of falling off)

Thus, the mean is:

extMean=30imes0.2=6 ext{Mean} = 30 imes 0.2 = 6

So, the mean number of times Patrick falls off in a day is 6.

Step 2

Find the variance of the number of times he falls off in a day.

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Answer

The variance of a binomial distribution can be calculated using the formula:

extVariance=nimespimes(1p) ext{Variance} = n imes p imes (1 - p)

Substituting the values:

  • n=30n = 30
  • p=0.2p = 0.2

We have:

extVariance=30imes0.2imes(10.2) ext{Variance} = 30 imes 0.2 imes (1 - 0.2)

Calculating:

extVariance=30imes0.2imes0.8=4.8 ext{Variance} = 30 imes 0.2 imes 0.8 = 4.8

Therefore, the variance is 4.8.

Step 3

Find the probability that, on a particular day, he falls off exactly 10 times.

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Answer

To find the probability of exactly kk successes in a binomial distribution, we use:

P(X = k) = inom{n}{k} p^k (1 - p)^{n - k}

For this problem, with n=30n = 30, k=10k = 10, and p=0.2p = 0.2, we have:

P(X = 10) = inom{30}{10} (0.2)^{10} (0.8)^{20}

Calculating this yields:

P(X=10)ext=approximately0.0355P(X = 10) ext{ = approximately } 0.0355

Thus, the probability of falling off exactly 10 times is approximately 0.0355.

Step 4

Find the probability that, on a particular day, he falls off 5 or more times.

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Answer

To find the probability of falling off 5 or more times, we can derive it from the cumulative probability:

P(Xextext5)=1P(X<5)P(X ext{ } ext{≥ 5}) = 1 - P(X < 5)

which is equivalent to:

P(Xextext5)=1[P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)]P(X ext{ } ext{≥ 5}) = 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)]

Using the binomial formula to calculate these probabilities:

After calculating, we find:

P(Xextext5)ext=approximately0.745P(X ext{ } ext{≥ 5}) ext{ = approximately } 0.745

Therefore, the probability of falling off 5 or more times is approximately 0.745.

Step 5

Calculate the probability that he will fall off his skateboard at least 5 times on each of the 5 days.

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Answer

The probability he falls off at least 5 times on any single day is already calculated as:

P(Xextext5)ext=0.745P(X ext{ } ext{≥ 5}) ext{ = } 0.745

For 5 consecutive days, assuming that each day's probability is independent, we raise this probability to the power of 5:

P(extAtleast5timeson5days)=(0.745)5P( ext{At least 5 times on 5 days}) = (0.745)^5

Calculating this gives:

(0.745)5ext=approximately0.229(0.745)^5 ext{ = approximately } 0.229

Thus, the probability of falling off at least 5 times on each of the 5 days is approximately 0.229.

Step 6

Explain why it may be unrealistic to use the same value of 0.2 for the probability of falling off for all 5 days.

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Answer

Using a constant probability of 0.2 assumes that Patrick's skill level and ability remain unchanged throughout the 5 days. However, as he practices, he is likely to improve, and his likelihood of falling off may decrease. Factors such as fatigue or external conditions could also affect his performance, leading to variability in the probability of falling off on different days.

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