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The random variable X has a normal distribution with mean 20 and standard deviation 4 - Edexcel - A-Level Maths: Statistics - Question 6 - 2007 - Paper 2

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The random variable X has a normal distribution with mean 20 and standard deviation 4. (a) Find P(X > 25). (b) Find the value of d such that P(20 < X < d) = 0.4641... show full transcript

Worked Solution & Example Answer:The random variable X has a normal distribution with mean 20 and standard deviation 4 - Edexcel - A-Level Maths: Statistics - Question 6 - 2007 - Paper 2

Step 1

Find P(X > 25)

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Answer

To find P(X > 25), we start by standardizing the variable X using the Z-score formula:

Z=X−μσZ = \frac{X - \mu}{\sigma}

where μ\mu = 20 (mean) and σ\sigma = 4 (standard deviation).

Substituting the values:

Z=25−204=54=1.25Z = \frac{25 - 20}{4} = \frac{5}{4} = 1.25

Using the standard normal distribution table or a calculator, we find:

P(Z>1.25)=1−P(Z≤1.25)≈1−0.8944=0.1056P(Z > 1.25) = 1 - P(Z \leq 1.25) \approx 1 - 0.8944 = 0.1056

Thus,

P(X>25)≈0.1056.P(X > 25) \approx 0.1056.

Step 2

Find the value of d such that P(20 < X < d) = 0.4641

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Answer

We want to find the value of dd such that:

P(20<X<d)=0.4641.P(20 < X < d) = 0.4641.

First, we know that:

P(X<d)=P(X<20)+P(20<X<d)=0.50+0.4641=0.9641.P(X < d) = P(X < 20) + P(20 < X < d) = 0.50 + 0.4641 = 0.9641.

Now, we standardize the upper bound dd:

P(Z<z)=0.9641P(Z < z) = 0.9641

Using the Z-table, we find that:

z≈1.80.z \approx 1.80.

Now, we can use the Z-score formula to find dd:

1.80=d−2041.80 = \frac{d - 20}{4}

Solving for dd, we have:

d=1.80×4+20=27.2.d = 1.80 \times 4 + 20 = 27.2.

Thus, the value of dd is:

d=27.2.d = 27.2.

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