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The measure of intelligence, IQ, of a group of students is assumed to be Normally distributed with mean 100 and standard deviation 15 - Edexcel - A-Level Maths Statistics - Question 7 - 2007 - Paper 1

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The measure of intelligence, IQ, of a group of students is assumed to be Normally distributed with mean 100 and standard deviation 15. (a) Find the probability that... show full transcript

Worked Solution & Example Answer:The measure of intelligence, IQ, of a group of students is assumed to be Normally distributed with mean 100 and standard deviation 15 - Edexcel - A-Level Maths Statistics - Question 7 - 2007 - Paper 1

Step 1

Find the probability that a student selected at random has an IQ less than 91.

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Answer

To find this probability, we need to standardize the value using the formula:

Z=XμσZ = \frac{X - \mu}{\sigma}

where:

  • X=91X = 91
  • μ=100\mu = 100
  • σ=15\sigma = 15

First, we calculate the Z-score:

Z=9110015=915=0.6Z = \frac{91 - 100}{15} = \frac{-9}{15} = -0.6

Using the Z-table, we find:

P(Z<0.6)=0.2743P(Z < -0.6) = 0.2743

Thus, the probability that a student has an IQ less than 91 is approximately 0.2743.

Step 2

Find, to the nearest integer, the value of k.

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Answer

We know that the probability of selecting a student with an IQ of at least 100+k100 + k is given as 0.2090:

P(X>100+k)=0.2090P(X > 100 + k) = 0.2090

This can also be expressed as:

P(X<100+k)=10.2090=0.7910P(X < 100 + k) = 1 - 0.2090 = 0.7910

Now standardizing this using the Z-score formula:

Z=(100+k)10015=k15Z = \frac{(100 + k) - 100}{15} = \frac{k}{15}

Using a Z-table (or calculator), we find that the Z value for 0.7910 is approximately 0.81:

k15=0.81\frac{k}{15} = 0.81

Now, solving for k gives:

k=0.81×15=12.15k = 0.81 \times 15 = 12.15

Rounding to the nearest integer, we find that k = 12.

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