It is known from the past experience that the number of telephone calls made daily in a certain community between 3 p.m. and 4 p.m. has a mean of 352 and standard deviation of 31. What percentage of the time will there be more than 400 telephone calls made in the community between 3 p.m. to 4 p.m.? (Use: $P(0 ≤ Z ≤ 1.55) = 0.4394$)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 6.06%
Given $μ = 352, σ = 31$ and $X = 400$, then
$Z=\frac{X-μ}{σ}=\frac{400-352}{31}⇒ Z = 1.55$
So, the required probability = $P(X > 400) = P(Z >1.55)$
Since area under the normal curve right to mean 0 is 0.5
∴ The required probability = $P(Z >1.55)$
$= 0.5-P(0 ≤ Z ≤ 1.55)$
$= 0.5-0.4394 = 0.0606$
Hence, the required percentage of the time = 6.06%