Question

The vendor at Citi Field offers a health pack consisting of apples and oranges. The weight, X, of an apple has a normal distribution with a mean of 9 ounces and a standard deviation of 0.6 ounces. Independent of this, the weight, Y, of an orange has a normal distribution with a mean of 7 ounces and a standard deviation of 0.4 ounces. Suppose the health pack has a random selection of 4 apples with weights

X₁, X₂, X₃, X₄

and 3 oranges with weights

Y₁, Y₂, Y₃

. . Let Xsum be the sum of the apple weights in ounces and let Ysum be the sum of the orange weights in ounces. W = Xsum + Ysum is the random variable representing the total weight of the health pack.


a) What is the probability that Xsum > 38?  

b) What is the probability that Ysum > 22?  

c) What is the expected value of Xsum?  

d) What is the standard deviation of Xsum?  

e) What is the variance of the random variable W?  

f) What is the expected value of W?  

g) What is the standard deviation of W?  

h) What is the probability that W > 59 ounces?  

i) i. If 100 health packs are sold what is the expected number sold which weigh more than 59 ounces?

Answer 1

The random variable X, Y and Z follows normal distribution,

XN(u,2)= i = 9, o2 = 0.62 0.36

0.16 = 7, o2 - 0.42 Y~N(,2

a)

38 P(Xsum 38) P(Xsum> P(Xsum > 9.5) 4

Since X follow normal distribution, probability is obtained by calculating z score,

9.5-9 9.5- 1.6667 0.6//4

P(Xsum > 38) P(Xsum 9.5) P(z> 1.6667) 1- P(z<1.6667)

Using the standard normal distribution table,

$P(Xsum>38)=1-0.95221=0.04779$

b)

22 P(Y sum> 7.3333) P(Y sum > 22) P(Ysum

Since X follow normal distribution, probability is obtained by calculating z score,

7.3333 7.3333 7 1.443376 0.4//3

P(z1.443376) 1 P(z < P(Y 22) P(Ys 1.443376) 7.3333 sum> sum

Using the standard normal distribution table,

$P(Ysum>22)=1-0.9255=0.0745$

c)

Using the property, E[aX]=aE[X]

$\text{Expected Value, }E[Xsum]=9\times 4 = 36$

d)

Using the property, Var[aX]=a²Var[X]

Variance, VarXsum] =42 x 0.36 5.76

$\text{Std. Dev, }=\sqrt{Var[Xsum]}=\sqrt{4^2\times 0.36 }= 2.4$

e)

Variance, VarXsum] =42 x 0.36 5.76

f)

Using the property, E[X+Y]=E[X]+E[Y]

$\text{Expected Value, }E[W]=E[X]+E[Y]=9\times 4+7\times 3=57$

g)

Using the property, Var[X+Y]=Var[X]+Var[Y]

$\text{Variance, }Var[Xsum+Ysum]=4^2\times 0.36+3^2\times 0.16=7.2$

$\text{Std. Dev, }=\sqrt{Var[W]}=\sqrt{7.2}= 2.6833$

h)

Since W follow normal distribution, probability is obtained by calculating z score,

$z=\frac{59-\mu}{\sigma}=\frac{59-57}{2.6833}=0.7454$

$P(W>59)=P(z>0.7454)=1-P(z<0.7454)$

Using the standard normal distribution table,

P(W59)1-0.7720 0.2546

i)

$\text{Expected Value, }E[W>59]=100\times 0.2546 = 25.46\approx 25$