Question

Assume that 5 % of the apples weigh less than 150 and 20 % of the apples weigh more than 225 . If the distribution of the weight of the apples is normal, find the mean and standard deviation of the distribution.

Answer 1

Consider two random variables X and Y with

σX=2σY=3Var[4X−5Y]=481.

What is the correlation between X and Y? That is, calculate ρXY.

Solution:

First, by definition of the variance of two random variables, we have

Var[4X−5Y]=(4)2⋅σ2X+(−5)2⋅σ2Y+(2)(4)(−5)σXY

Thus,

481=(4)2⋅22+(−5)2⋅32+(2)(4)(−5)σXY

Which gives

σXY=−4.8

Then, finally,

ρXY=σXYσX⋅σY=−4.82⋅3=−0.80

Answer 2

Answer:

The mean is μ = 200 and the standard deviation is σ = 30.

Step-by-step explanation:

Consider a random variable X representing the weight of apples.

It is mentioned that $X\sim N(\mu, \sigma^{2})$.

The values of μ and σ are unknown.

It is provided that,

$P(X < x_{1}) = 0.05\\\\P(X > x_{2}) = 0.20$

Rewrite the probability statements in terms of z-scores as follows:

$P(Z < z_{1})=0.05\\\\P(Z > z_{2})=0.20$

The corresponding z-scores are:

z1:

$P(Z < z_{1}) = 0.05\\\\z_{1}=Z_{0.05}=-1.65$

z2:

$P(Z > z_{2}) = 0.20\\\\1-P(Z < z_{2}) = 0.20\\\\P(Z < z_{2}) = 0.80\\\\z_{2}=Z_{0.80}=0.84$

**Use the z-table for the z-scores.

Now, using the value of $z_{1}$ form an equation in terms of $\mu$ and $\sigma$ as follows:

$z_{1}=\frac{x_{1}-\mu}{\sigma}\\\\-1.65=\frac{150-\mu}{\sigma}\\\\\mu-1.65\sigma=150\ ...(i)$

Again, using the value of $z_{2}$ form an equation in terms of $\mu$ and $\sigma$ as follows:

$z_{2}=\frac{x_{1}-\mu}{\sigma}\\\\0.84=\frac{225-\mu}{\sigma}\\\\\mu+0.84\sigma=225\ ...(ii)$

Solve equations (i) and (ii) simultaneously as follows:

$\mu-1.65\sigma=150\ \ \ \ \ (Multiply\ by\ 0.84)\\\\\mu+0.84\sigma=225\ \ \ \ \ (Multiply\ by\ 1.65)\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\\\0.84\mu-1.386\sigma=126\\\\1.65\mu+1.386\sigma=371.25\\\\ (Add\ the\ two\ equations)\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\\\2.49\mu=497.25\\\\\mu=199.6988\\\\\mu\approx 200$

Substitute the value of $\mu$ in equation (i) and solve for $\sigma$ as follows:

$\mu-1.65\sigma=150\\\\200-1.65\sigma=150\\\\1.65\sigma=200-150\\\\\sigma=\frac{50}{1.65}\\\\\sigma=30.30303\\\\\sigma\approx 30$

Thus, the mean and standard deviation for the distribution of the weight of apples are μ = 200 and σ = 30 respectively.