A-Level Edexcel Maths Statistics Normal Distribution: Past records show that the times

Past records show that the times, in seconds, taken to run 100 m by children at a school can be modelled by a normal distribution with a mean of 16.12 and a standard deviation of 1.60 A child from the school is selected at random - Edexcel - A-Level Maths Statistics - Question 4 - 2011 - Paper 2

Question 4

Past records show that the times, in seconds, taken to run 100 m by children at a school can be modelled by a normal distribution with a mean of 16.12 and a standard... show full transcript

Worked Solution & Example Answer:Past records show that the times, in seconds, taken to run 100 m by children at a school can be modelled by a normal distribution with a mean of 16.12 and a standard deviation of 1.60 A child from the school is selected at random - Edexcel - A-Level Maths Statistics - Question 4 - 2011 - Paper 2

Step 1

Find the probability that this child runs 100 m in less than 15 s.

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Answer

To solve this, we need to standardize the value 15 using the z-score formula:

$z = \frac{x - \mu}{\sigma}$

Where:

$x = 15$ (the value we are looking at)
$\mu = 16.12$ (mean)
$\sigma = 1.60$ (standard deviation)

Calculating the z-score:

$z = \frac{15 - 16.12}{1.60} = \frac{-1.12}{1.60} \approx -0.70$

Next, we refer to the standard normal distribution table to find the probability:

$P(Z < -0.70) \approx 0.2420$

Thus, the probability that the child runs 100 m in less than 15 s is approximately 0.242.

Step 2

Estimate, to 2 decimal places, the slowest time taken to run 100 m for which a child will be awarded a certificate.

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Answer

To find the cut-off time for the fastest 30% of the children, we need to find the z-score corresponding to the 70th percentile (since the slowest 30% means we are looking at the top 70%).

From the z-table, the z-score for the 70th percentile is approximately 0.524. We can now use the z-score formula again:

$P(T < t) = P \left( z = \frac{t - \mu}{\sigma} \right)$

Rearranging for $t$ :

$t = \mu + z \sigma$

Substituting in the known values:

= 16.9584 $$ Rounding to two decimal places gives: $$ t \approx 16.96 $$ Therefore, the slowest time taken to run 100 m for which a child will be awarded a certificate is approximately 16.96 seconds.

Past records show that the times, in seconds, taken to run 100 m by children at a school can be modelled by a normal distribution with a mean of 16.12 and a standard deviation of 1.60 A child from the school is selected at random - Edexcel - A-Level Maths Statistics - Question 4 - 2011 - Paper 2

Find the probability that this child runs 100 m in less than 15 s.

Estimate, to 2 decimal places, the slowest time taken to run 100 m for which a child will be awarded a certificate.

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