The number of telephone calls made daily in a certain community between 8 P.M. and 9 P.M. has a mean of 352 and standard deviation of 31. What percentage of time will there be more than 400 telephone calls made in this community between 8 P.M. to 9 P.M.?
[Given that: P( \( 0\le Z\le 1.55 \) ) = 0.4394, where \( Z \) is the standard normal variate]
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 6.06 %
Let the number of calls X be normally distributed with \( \mu=352 \), \( \sigma=31 \). The standard normal variate for 400 calls is
\( \displaystyle Z=\frac{X-\mu}{\sigma}=\frac{400-352}{31} \)
\( \displaystyle =\frac{48}{31}\approx 1.55 \)
By symmetry, half the area (0.5) lies to the right of Z = 0:
\( \displaystyle P(X\gt 400)=P(Z\gt 1.55) \)
\( \displaystyle =0.5-0.4394 \)
\( \displaystyle =0.0606 \)
So the percentage of time is 6.06%.
Tip: The table value P(0 ≤ Z ≤ z) is the area from the mean; for “more than”, subtract it from 0.5, not from 1.