Question

The Yankee Stadium food menu includes dirty fries and loaded fries. Let D be the weight, in ounces, of a randomly selected order of dirty fries. Let L be the weight, in ounces, of a randomly selected order of loaded fries. Suppose D is normally distributed with a mean weight of 5.2 ournces and a standard deviation of 0.4 ounces. Suppose L is normally distributed with a mean weight of 5.4 ounces and a standard deviation of 0.5 ounces. L and D are independent. a) Calculate the expected value of L D. (in ounces) b) Calculate the standard deviation of L +D. (in ounces) c) What is the probability that an order of loaded fries weighs less than 5 ounces?? d) What is the probability that an order of loaded fries weighs between 5 and 6 ounces? e) What is the 95th percentile of D? f) What is the probability that L D is greater than 11 ounces? g) Suppose we purchase an order of loaded fries and an order of dirty fries. What is the probability that each of the orders weighs more than 5 ounces? (L>5 and D>5) h)h. What is the probability that D is greater than L?

Answer 1

Given:

Let L be the weight, in ounces, of a randomly selected order of loaded fries.

$L\sim N(\mu_{L} =5.4,\sigma_{L} =0.5)$;

where $\mu_{L} =E(L)$ is the mean of the random variable L and $\sigma _{L}$ is the standard deviation of the random variable L.

Let D be the weight, in ounces, of a randomly selected order of dirty fries.

$D\sim N(\mu_{D} =5.2,\sigma_{D} =0.4)$

where $\mu _{D}=E(D)$ is the mean of the random variable D and $\sigma _{D}$ is the standard deviation of the random variable D.

In the below solution, let Z be a standard normal variate or $Z\sim N(0,1)$

a) Expected value of L+D is $E(L+D)=E(L)+E(D)$ , since expectations are linear.

Therefore, $E(L+D)=5.4+5.2=10.6$ .

Hence, Expected value of L+D is 10.6 ounces.

b) Variance of L+D is given by $V(L+D)=V(L)+V(D)+2cov(L,D)=V(L)+V(D)$ , since $cov(L,D)=E(LD)-E(L)E(D)=0$ , since L and D are independent.

Therefore, $V(L+D)=0.4^{2}+0.5^{2}=0.41$ , since variance is the square of standard deviation.

$\sigma _{L+D}=\sqrt{0.41}=0.6403$

Standard Deviation of L+D is 0.6403 ounces.

c)  $P(L<5)=P(\frac{L-\mu _{L}}{\sigma _{L}}<\frac{5-5.4}{0.5})$ (standardizing L into the normal variate)

$P(Z<-0.8)=1-P(Z<0.8)=1-0.7881=0.2119$ ( by referring to the normal table values).

Therefore, the probability that an order of loaded fries weighs less than 5 ounces is 0.2119.

d) $P(5<L<6)=P(\frac{5-5.4}{0.5}<\frac{L-\mu _{L}}{\sigma _{L}}<\frac{6-5.4}{0.5})=P(-0.8<Z<1.2)$

$=P(Z<1.2)-P(Z<-0.8)=0.8849-0.2119=0.6730$,(by properties of normal distribution)

Therefore, the probability that an order of loaded fries weighs between 5 and 6 ounces is 0.6730.

e) $P(D<p)=0.95$.Here, p is the 95^th percentile of the random variable D.

$P(\frac{D-\mu _{D}}{\sigma _{D}}<\frac{p-5.2}{0.4})=0.95$

$P(Z<\frac{p-5.2}{0.4})=0.95$

The 95th percentile of Z is found to be 1.645. ( This can be checked from the normal table).

$\frac{p-5.2}{0.4}=1.645$

Therefore, p=(1.645*0.4)+5.2=5.858

The 95th percentile of D is 5.858.

f) Let U = L+D,

We know from the questions (a) and (b) that $\mu _{U}$=E(U)=10.6 and $\sigma _{U}=0.6403$

$P(U>11)=P(\frac{U-\mu _{U}}{\sigma _{U}}>\frac{11-10.6}{0.6403})=P(Z>0.625)$

$=1-P(Z\leq 0.625)=1-0.7340=0.266$

Therefore, the probability that L+D weighs greater than 11 ounces is 0.266.

g) $P(L>5 \: and \: D>5)=P(L>5\bigcap D>5)=P(L>5)*P(D>5)$ , since L and D are independent.

$P(L>5)=P(\frac{L-\mu _{L}}{\sigma _{L}}>\frac{5-5.4}{0.5})=P(Z>-0.8)=1-P(Z\leq -0.8)$

$1-(1-P(Z<0.8))=P(Z<0.8)=0.7881$

$P(D>5)=P(\frac{D-\mu _{D}}{\sigma _{D}}>\frac{5-5.2}{0.4})=P(Z>-0.5)=1-P(Z\leq -0.5)$

$1-(1-P(Z<0.5))=P(Z<0.5)=0.6915$

P(L>5 and D>5) = 0.7881*0.6915= 0.5449

Hence, the probability that each of the order weighs more than 5 ounces is 0.5449.

h) $P(D>L)=P(D-L>0)$

$E(D-L)=E(D)-E(L)=5.4-5.2=0.2$

$V(D-L)=V(D)&plus;V(L)-2*cov(D,L)=V(D)&plus;V(L)=0.4^{2}&plus;0.5^{2}$

$=0.41$,since $cov(L,D)=E(LD)-E(L)E(D)=0$ , since L and D are independent.

$\sigma _{D-L}=\sqrt{0.41}=0.6403$

$P(D-L>0)=P(Z>\frac{0-0.2}{0.6403})=P(Z>-0.3123)$

$=1-P(Z\leq 0.3123)=1-0.6217=0.3783$

Therefore, the probability that D is greater than L is 0.3783.