Chapter 18, Problem 54E

(a)

To determine

To find the probability that she averages more than $50$ pounds of milk a day.

Answer to Problem 54E

  $0.3085$ .

Explanation of Solution

In the question, Ayrshire cows average $47$ pounds of milk a day with a standard deviation of $6$ pounds of milk. For Jersey cows, the mean daily production is $43$ ponds with a standard deviation of $5$ pounds. And assume that Normal models describe milk production for these breeds. Thus, let us first find the z -score as:

  $\begin{array}{l}z=\frac{50-47}{6}\\ =0.5\end{array}$

And the probability that she averages more than $50$ pounds of milk a day is calculated as:

  $\begin{array}{l}P(z\ge 0.5)=normalcdf(0.5,E99,0,1)\\ =0.3085\end{array}$

Thus, the probability that that she averages more than $50$ pounds of milk a day is approximately $0.3085$ .

(b)

To determine

To find the probability that a randomly selected Ayrshire gives more milk than randomly selected Jersey.

Answer to Problem 54E

  $0.6957$ .

Explanation of Solution

In the question, Ayrshire cows average $47$ pounds of milk a day with a standard deviation of $6$ pounds of milk. For Jersey cows, the mean daily production is $43$ ponds with a standard deviation of $5$ pounds. And assume that Normal models describe milk production for these breeds. Thus, to find the probability we need to find the difference mean and standard deviation as:

  $\begin{array}{l}\mu =E(X-Y)\\ =47-43\\ =4\\ \sigma =SD(X-Y)\\ =\sqrt{Var(X)+Var(Y)}\\ =\sqrt{36+25}\\ =7.8102\end{array}$

Then the z -score will be as:

  $\begin{array}{l}z=\frac{0-4}{7.8102}\\ =-0.5122\end{array}$

Thus, the probability that a randomly selected Ayrshire gives more milk than randomly selected Jersey is calculated as:

  $\begin{array}{l}P(z\ge -0.5122)=normalcdf(-0.5122,E99,0,1)\\ =0.6957\end{array}$

Thus, the probability that a randomly selected Ayrshire gives more milk than randomly selected Jersey is approximately $0.6957$ .

(c)

To determine

To find the probability that the average production for this small herd exceeds $45$ pounds of milk in a day.

Answer to Problem 54E

  $0.03681$ .

Explanation of Solution

In the question, Ayrshire cows average $47$ pounds of milk a day with a standard deviation of $6$ pounds of milk. For Jersey cows, the mean daily production is $43$ ponds with a standard deviation of $5$ pounds. And assume that Normal models describe milk production for these breeds. Thus, we have,

  $\begin{array}{l}{\mu }_{\overline{y}}=43\\ {\sigma }_{\overline{y}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{5}{\sqrt{20}}\\ =1.118\end{array}$

Then the z -score will be as:

  $\begin{array}{l}z=\frac{45-43}{1.118}\\ =1.789\end{array}$

Thus, the probability that the average production for this small herd exceeds $45$ pounds of milk in a day is calculated as:

  $\begin{array}{l}P(z\ge 1.789)=normalcdf(1.789,E99,0,1)\\ =0.03681\end{array}$

Thus, the probability that the average production for this small herd exceeds $45$ pounds of milk in a day is approximately $0.03681$ .

(d)

To determine

To find out what is the probability that his herd average is at least $5$ pounds higher than the average for part (c)’s Jersey herd.

Answer to Problem 54E

  $0.3249$ .

Explanation of Solution

In the question, Ayrshire cows average $47$ pounds of milk a day with a standard deviation of $6$ pounds of milk. For Jersey cows, the mean daily production is $43$ ponds with a standard deviation of $5$ pounds. And assume that Normal models describe milk production for these breeds. Thus, we have,

  $\begin{array}{l}\mu ={\mu }_{x-y}\\ =47-43\\ =4\\ \sigma =\sqrt{Var(X)+Var(Y)}\\ =2.2023\end{array}$

Thus, the z -score will be then,

  $\begin{array}{l}z=\frac{5-4}{2.2023}\\ =0.4541\end{array}$

And the probability that his herd average is at least $5$ pounds higher than the average for part (c)’s Jersey herd is calculated as:

  $\begin{array}{l}P(z\ge 0.4541)=normalcdf(0.4541,E99,0,1)\\ =0.3249\end{array}$

So, the probability that his herd average is at least $5$ pounds higher than the average for part (c)’s Jersey herd is approximately $0.3249$ .