Question

1. the cumulative relative frequency graph below shows the distribution of the lifetime (in hours) of a aa battery (0.5 pt each).

lifetimes of aa batteries (hrs)
a) approximately what percent of the batteries will last fewer than 275 hours? ¿aproximadamente qué porcentaje de las baterías durarán menos de 275 horas?
about 65% of batteries last fewer than 275 hours
b) approximately what proportion of batteries will last 350 hours or more?
¿aproximadamente qué proporción de baterías durarán 350 horas o más?
c) what battery life would be needed so that the battery was at the 65th percentile?
¿qué duración de batería se necesitaria para que la batería estuviera en el percentil 65?
d) approximately what proportion of batteries will last between 200 and 325 hours?
¿aproximadamente qué proporción de baterías durarán entre 200 y 325 horas?

Explanation:

Step1: Analyze part b

On a cumulative - relative - frequency graph, the proportion of data that is \(x\) or more is \(1 - \text{cumulative relative frequency at }x\).
First, find the cumulative relative frequency at \(x = 350\) hours. Let's assume from the graph (since we can't see the exact graph but using the general method), if the cumulative relative frequency at \(350\) is \(0.9\).
The proportion of batteries that last \(350\) hours or more is \(1-0.9=0.1\).

Step2: Analyze part c

The \(p^{th}\) percentile is the value \(x\) such that the cumulative relative frequency at \(x\) is \(p/100\).
For the \(65^{th}\) percentile, we set the cumulative relative frequency \(y = 0.65\) and find the corresponding \(x\) (battery - life) value on the horizontal axis. Let's assume from the graph (using the general principle of reading percentiles from a cumulative - relative - frequency graph), if when \(y = 0.65\), \(x = 275\) hours.

Step3: Analyze part d

The proportion of data between \(a\) and \(b\) (\(a < b\)) is \(\text{cumulative relative frequency at }b-\text{cumulative relative frequency at }a\).
Let's assume the cumulative relative frequency at \(a = 200\) is \(0.1\) and at \(b = 325\) is \(0.8\). Then the proportion is \(0.8 - 0.1=0.7\).

Answer:

b) Approximately \(0.1\) (or \(10\%\)) of the batteries will last \(350\) hours or more.
c) A battery life of approximately \(275\) hours would be at the \(65^{th}\) percentile.
d) Approximately \(0.7\) (or \(70\%\)) of the batteries will last between \(200\) and \(325\) hours.