Question

The accompanying frequency distribution of fracture strength (MPa) observations for ceramic bars fired in aparticular kiln appeared in the article “Evaluating TunnelKiln Performance” (Amer. Ceramic Soc. Bull., Aug.1997: 59–63).

Class81-<83 83-<85 85-<87 87-<89 89-<91

Frequency6 7 17 30 43

Class91-<93 93-<95 95-<97 97-<99

Frequency28 22 13 3

1. Construct a histogram based on relative frequencies, and comment on any interesting features.
2. What proportion of the strength observations are at least 85? Less than 95?

c. Roughly what proportion of the observations are less than 90?

Short answer

1. The histogram is shown below:

The distribution is symmetric and does not contain any outliers.

b. The proportion of the strength observations that are at least 85 is 0.923.The proportion of the strength observations are less than 95 is 0.905.

c. The proportion of the observations that are less than 90 is 0.4822.

Step-by-step solution

Given information

Frequency distribution of fracture strength (MPa) observations for ceramic bars fired in a particular kiln is provided.

Construct a histogram and state the features

a.

The relative frequency is computed as,

\(\begin{aligned}{\rm{Relative frequency }} &= \frac{{{\rm{Frequency}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{observations}}}}\\{R_i} &= \frac{{{f_i}}}{n}\end{aligned}\)

The table representing the relative frequency is computed as,

Class (i)	Frequency\(\left( {{f_i}} \right)\)	Relative Frequency\(\left( {{R_i}} \right)\)
81-<83	6	0.036
83-<85	7	0.041
85-<87	17	0.101
87-<89	30	0.178
89-<91	43	0.254
91-<93	28	0.166
93-<95	22	0.130
95-<97	13	0.077
97-<99	3	0.018
Total	169	1

Steps to construct a histogram are,

1) Determine the frequency or the relative frequency.

2) Mark the class boundaries on the horizontal axis.

3) Draw a rectangle on the horizontal axis corresponding to the frequency or relative frequency.

The histogram is represented as,

The features that can be observed from the above histogram are,

1)The distribution is symmetric as the histogram forms a almost perfect u-shaped mounted curve; this implies that mean, median and mode are equal.

2) There are no outliers present in the data.

Calculate the proportions

b.

Let x represents the fracture strength observations.

Referring to the relative frequencies computed in part a,

The proportion of the strength observations are at least 85 is computed as,

\(\begin{aligned}{R_{\left( {x \ge 85} \right)}} &= 1 - {R_{\left( {x < 85} \right)}}\\ &= 1 - \left( {0.036 + 0.041} \right)\\ &= 0.923\end{aligned}\)

Thus, the proportion of the strength observations are at least 85 is 0.923.

The proportion of the strength observations are less than 95 is computed as,

\(\begin{aligned}{R_{\left( {x < 95} \right)}} &= 1 - {R_{\left( {x \ge 95} \right)}}\\ &= 1 - \left( {0.077 + 0.018} \right)\\ &= 0.905\end{aligned}\)

Thus, the proportion of the strength observations are less than 95 is 0.905.

Calculate the proportion

c.

Referring to the relative frequencies computed in part a,

The proportion of the observations that are less than 90 is roughly the half of proportion of class 89-91 and the proportions that less than 89 (for the provided data).

Mathematically,

\(\begin{aligned}{R_{\left( {x < 90} \right)}} &= 0.036 + 0.041 + 0.101 + 0.178 + \frac{1}{2}\left( {0.254} \right)\\ &= 0.4822\end{aligned}\)

Thus, the proportion of the observations that are less than 90 is 0.4822.