The random variable XX has distribution N(μ,σ2)\mathrm{N}(\mu, \sigma^2).

  1. (a)

    Given that P(X>k)=0.45\mathrm{P}(X > k) = 0.45, find the value of P(X>2μ−k)\mathrm{P}(X > 2\mu - k).

  2. (b)

    It is given that μ=4\mu = 4 and P(1<X<4)=0.2475\mathrm{P}(1 < X < 4) = 0.2475.

    Find the value of σ\sigma.

  3. (c)

    Draw a sketch to show the distribution of XX for xx between −11-11 and 1919.

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Worked solution

9(a)

1. Identify the symmetrical relationship between $2\mu - k$ and $k$ about the mean $\mu$.

The normal distribution curve is symmetrical about x=μx = \mu.
The distance from 2μ−k2\mu - k to the mean μ\mu is μ−(2μ−k)=k−μ\mu - (2\mu - k) = k - \mu, which is exactly the same as the distance from μ\mu to kk. Thus, 2μ−k2\mu - k is the reflection of kk across the mean.

2. Apply the symmetry property and the complement rule to calculate the probability.

By symmetry, the probability to the right of 2μ−k2\mu - k is equal to the probability to the left of kk:
P(X>2μ−k)=P(X<k)\mathrm{P}(X > 2\mu - k) = \mathrm{P}(X < k)
Using the complement rule:
P(X<k)=1−P(X>k)\mathrm{P}(X < k) = 1 - \mathrm{P}(X > k)
1−0.45=0.551 - 0.45 = 0.55

Answer

0.550.55

9(b)

1. Find the probability that $X < 1$.

Since μ=4\mu = 4, the distribution is symmetric about X=4X = 4, so P(X<4)=0.5\mathrm{P}(X < 4) = 0.5.
P(X<1)=P(X<4)−P(1<X<4)\mathrm{P}(X < 1) = \mathrm{P}(X < 4) - \mathrm{P}(1 < X < 4)
P(X<1)=0.5−0.2475=0.2525\mathrm{P}(X < 1) = 0.5 - 0.2475 = 0.2525

2. Standardise the value and solve for $\sigma$ using the inverse normal distribution.

Standardising using Z=X−μσZ = \frac{X - \mu}{\sigma}:
P(Z<1−4σ)=0.2525\mathrm{P}\left(Z < \frac{1 - 4}{\sigma}\right) = 0.2525
Using the inverse normal function on a graphing calculator:
−3σ≈−0.666665\frac{-3}{\sigma} \approx -0.666665
σ=−3−0.666665≈4.50\sigma = \frac{-3}{-0.666665} \approx 4.50

Answer

4.504.50

9(c)

1. Calculate the standard deviation $\sigma$ using information from part (b).

P(1<X<4)=0.2475\mathrm{P}(1 < X < 4) = 0.2475
Since the normal curve is symmetrical about the mean μ=4\mu = 4, P(X<4)=0.5\mathrm{P}(X < 4) = 0.5.
P(X<1)=0.5−0.2475=0.2525\mathrm{P}(X < 1) = 0.5 - 0.2475 = 0.2525
Standardising the value:
P(Z<1−4σ)=0.2525\mathrm{P}\left(Z < \frac{1 - 4}{\sigma}\right) = 0.2525
−3σ=Φ−1(0.2525)≈−0.66667\frac{-3}{\sigma} = \Phi^{-1}(0.2525) \approx -0.66667
σ=4.5\sigma = 4.5

2. Determine the features of the sketch relative to the mean and standard deviation.

The normal curve is symmetrical and peaks at the mean, x=4x = 4.
The domain boundaries −11-11 and 1919 are equidistant from the mean:
4−(−11)=15and19−4=154 - (-11) = 15 \quad \text{and} \quad 19 - 4 = 15
Since 15=154.5σ≈3.33σ15 = \frac{15}{4.5}\sigma \approx 3.33\sigma, the required interval [−11,19][-11, 19] covers more than 33 standard deviations on either side of the mean. Therefore, the curve should practically taper off to the xx-axis at these boundaries.

3. Describe the required sketch.

  • Draw a horizontal xx-axis.
  • Sketch a bell-shaped curve strictly above the xx-axis.
  • Mark the maximum point on the curve at the line of symmetry x=4x = 4.
  • Extend the tails so they taper off and closely approach the xx-axis at the labeled endpoints x=−11x = -11 and x=19x = 19.
Answer

A symmetrical bell-shaped curve strictly above the xx-axis, peaking at x=4x = 4, and tapering off to approach the xx-axis near the endpoints x=−11x = -11 and x=19x = 19.

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