Question

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Answer 1

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=(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=(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*100 \\ =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=(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*100 \\ =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=(1000-1123)/(84) \\ =-1.464285714 \\ z_2=(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