$$\begin{array} { | l | l | }
\hline \text { Length (inch) } & \text { Frequency } \
\hline 0.1 & 2 \
\hline 0.2 & 3 \
\hline 0.3 & 3 \
\hline 0.4 & 5 \
\hline 0.5 & 4 \
\hline 0.6 & 2 \
\hline 0.7 & 2 \
\hline
\end{array}$$ Adele is studying a new species of insect that emits an odor perceived to be pleasant to humans. The known population consists of 21 insects. Adele has measured the length of each one.
An insect that is 0.7 inches long would have a deviation score of:
A) 0
B) 0.1
C) 0.2
D) 0.3

Question

$$\begin{array} { | l | l | } 
\hline \text { Length (inch) } & \text { Frequency } \
\hline 0.1 & 2 \
\hline 0.2 & 3 \
\hline 0.3 & 3 \
\hline 0.4 & 5 \
\hline 0.5 & 4 \
\hline 0.6 & 2 \
\hline 0.7 & 2 \
\hline
\end{array}$$ Adele is studying a new species of insect that emits an odor perceived to be pleasant to humans. The known population consists of 21 insects. Adele has measured the length of each one.
An insect that is 0.7 inches long would have a deviation score of:
A) 0
B) 0.1
C) 0.2
D) 0.3

Quizplus · Accepted Answer

The formula for calculating deviation score is:
Deviation Score = (Value of Data Point - Mean) / Standard Deviation

To find the deviation score for an insect that is 0.7 inches long, we first need to find the mean and standard deviation of the length distribution.

Mean = (0.1x2 + 0.2x3 + 0.3x3 + 0.4x5 + 0.5x4 + 0.6x2 + 0.7x2) / 21 = 0.391

To find the standard deviation, we first need to find the variance:
Variance = ((0.1-0.391)^2x2 + (0.2-0.391)^2x3 + (0.3-0.391)^2x3 + (0.4-0.391)^2x5 + (0.5-0.391)^2x4 + (0.6-0.391)^2x2 + (0.7-0.391)^2x2) / 21 = 0.042

Standard Deviation = sqrt(Variance) = 0.205

Now we can calculate the deviation score for an insect that is 0.7 inches long:
Deviation Score = (0.7 - 0.391) / 0.205 = 1.512

Rounding to the nearest tenth, we get a deviation score of 1.5. 
Therefore, the closest option is D) 0.3.

Adele Is Studying a New Species of Insect That Emits