Question

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Answer 1

Solution:

Given that a livestock company reports that the mean weight of a group of young steers is 1123 pounds with a standard deviation of 84 pounds, this implies that

$\begin{gathered} \mu=1123 \\ \sigma=84 \\ \end{gathered}$

A) Percentage of steers that weigh over 1300 pounds.

Step 1: Evaluate the z score value.

The z score value is expressed as

$\begin{gathered} z=\frac{x-\mu}{\sigma} \\ where \\ x\Rightarrow sample\text{ value} \\ \mu\Rightarrow mean\text{ value} \\ \sigma\Rightarrow standard\text{ deviation of the sample} \end{gathered}$

In this case, x equals 1300.

Thus, the z score value is evaluated as

$\begin{gathered} z=\frac{1300-1123}{84} \\ \Rightarrow z=2.107142857 \end{gathered}$

Step 2: Evaluate the probability that the steer would weigh over 1300 pounds.

Using the normal distribution table, we have

thus,

$\begin{gathered} Pr(z>1300)=0.0175525991947 \\ \end{gathered}$

The percentage of young steers that weigh over 1300 pounds is thus evaluated as

$\begin{gathered} 0.0175525991947\times100 \\ =1.7552991947\% \\ \approx1.8\%\text{ \lparen1 decimal place\rparen} \end{gathered}$

B) Percentage of steers that have weights under 1050 pounds.

Step 1: Evaluate the z score value.

Thus, we have

$\begin{gathered} z=\frac{1050-1123}{84} \\ =-0.869047619 \end{gathered}$

Step 2: Evaluate the probability that the steer would have weights under 1050 pounds.

Using the normal distribution table, we have

thus,

$Pr(z<1050)=0.192410542709$

Thus, the percentage of young steers that would have weight under 1050 pounds is evaluated as

$\begin{gathered} 0.192410542709\times100 \\ =19.2410542709\% \\ \approx19.2\%(\text{ 1 decimal place\rparen} \end{gathered}$

C) Percentage of young steers that weigh between 1000 and 1200 pounds.

Step 1: Evaluate the z score value.

$\begin{gathered} z_1=\frac{1000-1123}{84} \\ =-1.464285714 \\ z_2=\frac{1200-1123}{84} \\ =0.9166666667 \end{gathered}$

Step 2: Evaluate the probability that the steer would weigh between 1000 and 1200 pounds.

Using the normal distribution table, we have

thus, we have

$Pr(1000Thus, the percentage of young steers that weigh between 1000 and 1200 pounds is evaluated as[tex]\begin{gathered} 0.748783381404\times100 \\ =74.8783381404\% \\ \approx74.9\% \end{gathered}$