Question

We can expect that at least 75% of the bacteria to live between _____ and _____ minutes. (Enter x‾−ks in the first blank and x‾+ks in the second blank.)

a) x‾−0.674s and x‾+0.674s
b) x‾−1.645s and x‾+1.645s
c) x‾−1.282s and x‾+1.282s
d) x‾−2.576s and x‾+2.576s

Answer 1

We can expect that at least 75% of the bacteria to live between
`\( \overline{x} - 1.645s \)` and
`\( \overline{x} + 1.645s \)`.

Step-by-step explanation:

To determine the range within which at least 75% of the bacteria are expected to live, we use the properties of the standard normal distribution. The given values are associated with the z-score for a cumulative probability of 0.75. The z-scores represent the standard deviations from the mean
`(\( \overline{x} \))`.

The correct option is b)
`\( \overline{x} - 1.645s \)` and
`\( \overline{x} + 1.645s \)`. This corresponds to the z-score of approximately 1.645, which encompasses the central 75% of the distribution. It ensures that at least 75% of the bacteria fall within this range.

The other options do not match the correct z-score for a cumulative probability of 0.75. Options a), c), and d) correspond to different z-scores, which would result in incorrect estimates for the range within which at least 75% of the bacteria are expected to live.

In conclusion, option b) provides the correct z-scores and, therefore, the accurate range within which at least 75% of the bacteria are expected to live based on the standard normal distribution.