91影视

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

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,

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.

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.

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.