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
The Z-score for 300 s is
"\\begin{aligned}\nZ_{x} &=\\frac{x-\\bar x}{s} \\\\\nZ_{300} &=\\frac{300-225}{55} \\\\\n\n&=1.36\n\n\\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}\n\nZ_{200} &=\\frac{200-225}{55} \\\\\n\n&=-0.45\n\n\\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 \\%"
Comments
Leave a comment