Question

A telephone company has found that the lengths of its long distance telephone calls are normally distributed, with a mean of 225 s and a standard deviation of 55 s. What percent of its long distance calls are between 200 and 300 s

Answer 1

The Z-score for 300 s is

$\begin{aligned} Z_{x} &=\frac{x-\bar x}{s} \\ Z_{300} &=\frac{300-225}{55} \\ &=1.36 \end{aligned}$

Table indicates that 0.413(41.3%) of the data a normal distribution is between z=0 and z=1.36

The Z-score for 200 s is

$\begin{aligned} Z_{200} &=\frac{200-225}{55} \\ &=-0.45 \end{aligned}$

Table indicates that 0.174(17.4%) of the data is a normal distribution are between z=0 and z=-.45. Because the data are normally distributed, 17.4% of the data is also between z=0 and z=-.45. Thus percent of the long distance call are between 200 s and 300 s is $41.3 \%+17.4 \%=58.7 \%$