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Sagot :
Data given;
[tex]\begin{gathered} \mu=12hrs \\ \sigma=1hr \\ x=13\text{ and 14} \\ Z=\text{?} \end{gathered}[/tex]We have to find the Z score within both ranges, we use the formula below, substituting the given values.
[tex]\begin{gathered} Z=\frac{x-\mu}{\sigma} \\ Z_{13}=\frac{13-12}{1}=\frac{1}{1}=1 \\ Z_{14}=\frac{14-12}{1}=\frac{2}{1}=2 \end{gathered}[/tex]Since the Z-scores are known, we will get the probability from the Z-score table using the range (1
From the Z-score table, P(1
The number of phones that will have battery life in the 13 to 14 hours range will be;
0.13591 x 2800 = 380.548
Therefore, the number of phones that will have battery life in the 13 to 14 hours range is approximately 381 phones to the nearest whole number since a phone is whole and can not be a fraction.
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