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Question2X∼N(μ,152)2 X \sim N\left(\mu, 15^{2}\right). Given that P(X<90)=0.1P(X<90)=0.1, determine μ\mu. 3X∼N(40,σ2)3 X \sim N\left(40, \sigma^{2}\right). Given that P(X>50)=0.01P(X>50)=0.01, determine μ\mu. 4X∼N(μ,σ2)4 X \sim N\left(\mu, \sigma^{2}\right). Given that P(X<5)=0.05P(X<5)=0.05 and P(X>10)=0.01P(X>10)=0.01, determine μ\mu and σ\sigma.

Studdy Solution

STEP 1

1. 2X2X, 3X3X, and 4X4X are normally distributed random variables.
2. The properties of normal distribution and standard normal distribution (Z-scores) will be used.
3. The cumulative distribution function (CDF) and its inverse (quantile function) are applicable.

STEP 2

1. Solve for μ\mu in the equation 2X∼N(μ,152)2X \sim N(\mu, 15^2) given P(X<90)=0.1P(X<90)=0.1.
2. Solve for μ\mu in the equation 3X∼N(40,σ2)3X \sim N(40, \sigma^2) given P(X>50)=0.01P(X>50)=0.01.
3. Solve for μ\mu and σ\sigma in the equation 4X∼N(μ,σ2)4X \sim N(\mu, \sigma^2) given P(X<5)=0.05P(X<5)=0.05 and P(X>10)=0.01P(X>10)=0.01.

STEP 3

Given 2X∼N(μ,152)2X \sim N(\mu, 15^2), find the equivalent distribution for XX:
X∼N(μ2,(152)2) X \sim N\left(\frac{\mu}{2}, \left(\frac{15}{2}\right)^2\right)

STEP 4

Use the standard normal distribution to find the Z-score corresponding to P(X<90)=0.1P(X<90)=0.1:
Z=90−μ2152 Z = \frac{90 - \frac{\mu}{2}}{\frac{15}{2}}
From standard normal tables, Z≈−1.2816Z \approx -1.2816.

STEP 5

Solve for μ\mu:
−1.2816=90−μ2152 -1.2816 = \frac{90 - \frac{\mu}{2}}{\frac{15}{2}}
μ=2(90+1.2816×152) \mu = 2 \left(90 + 1.2816 \times \frac{15}{2}\right)
μ≈180+19.224=199.224 \mu \approx 180 + 19.224 = 199.224

STEP 6

Given 3X∼N(40,σ2)3X \sim N(40, \sigma^2), find the equivalent distribution for XX:
X∼N(403,(σ3)2) X \sim N\left(\frac{40}{3}, \left(\frac{\sigma}{3}\right)^2\right)

STEP 7

Use the standard normal distribution to find the Z-score corresponding to P(X>50)=0.01P(X>50)=0.01:
Z=50−403σ3 Z = \frac{50 - \frac{40}{3}}{\frac{\sigma}{3}}
From standard normal tables, Z≈2.3263Z \approx 2.3263.

STEP 8

Solve for σ\sigma:
2.3263=50−403σ3 2.3263 = \frac{50 - \frac{40}{3}}{\frac{\sigma}{3}}
σ=3×50−4032.3263 \sigma = 3 \times \frac{50 - \frac{40}{3}}{2.3263}
σ≈3×1103×2.3263 \sigma \approx 3 \times \frac{110}{3 \times 2.3263}
σ≈47.32 \sigma \approx 47.32

STEP 9

Given 4X∼N(μ,σ2)4X \sim N(\mu, \sigma^2), use the conditions P(X<5)=0.05P(X<5)=0.05 and P(X>10)=0.01P(X>10)=0.01 to set up two equations:
For P(X<5)=0.05P(X<5)=0.05:
Z1=5−μσ Z_1 = \frac{5 - \mu}{\sigma}
From standard normal tables, Z1≈−1.6449Z_1 \approx -1.6449.
For P(X>10)=0.01P(X>10)=0.01:
Z2=10−μσ Z_2 = \frac{10 - \mu}{\sigma}
From standard normal tables, Z2≈2.3263Z_2 \approx 2.3263.

STEP 10

Solve the system of equations for μ\mu and σ\sigma:
−1.6449=5−μσ -1.6449 = \frac{5 - \mu}{\sigma}
2.3263=10−μσ 2.3263 = \frac{10 - \mu}{\sigma}
Solve these equations simultaneously to find μ\mu and σ\sigma.

STEP 11

Subtract the first equation from the second:
2.3263+1.6449=10−5σ 2.3263 + 1.6449 = \frac{10 - 5}{\sigma}
σ=53.9712 \sigma = \frac{5}{3.9712}
σ≈1.259 \sigma \approx 1.259
Substitute σ\sigma back into one of the equations to find μ\mu:
−1.6449=5−μ1.259 -1.6449 = \frac{5 - \mu}{1.259}
μ=5+1.6449×1.259 \mu = 5 + 1.6449 \times 1.259
μ≈6.07 \mu \approx 6.07
The solutions are:
1. For 2X2X, μ≈199.224\mu \approx 199.224.
2. For 3X3X, σ≈47.32\sigma \approx 47.32.
3. For 4X4X, μ≈6.07\mu \approx 6.07 and σ≈1.259\sigma \approx 1.259.

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