klupko5HR

Answered

2022-12-05

Red Widget Strategies wants to improve its customer service by being more efficient in how phone calls are processed. Toward that end, Red Widget does a statistical analysis of 1,000 outbound phone calls. The length of the calls is normally distributed, with a population mean = 240 seconds, and a population standard deviation = 40 second.

a. What is the probability that a particular call lasted less than 180 seconds?

b. What is the probability that a particular call lasted between 180 and 300 seconds?

c. What is the probability that a call lasted between 110 and 180 seconds?

d. What is the probability that a call lasted more than 300 seconds?

a. What is the probability that a particular call lasted less than 180 seconds?

b. What is the probability that a particular call lasted between 180 and 300 seconds?

c. What is the probability that a call lasted between 110 and 180 seconds?

d. What is the probability that a call lasted more than 300 seconds?

Answer & Explanation

greffhc

Expert

2022-12-06Added 7 answers

Solution:

Given that,

$\mu =240\phantom{\rule{0ex}{0ex}}\sigma =40\phantom{\rule{0ex}{0ex}}N=1000$

$p(x180)\phantom{\rule{0ex}{0ex}}=p(x-\frac{\mu}{\sigma})(180-\frac{240}{40})\phantom{\rule{0ex}{0ex}}=p(z-\frac{60}{40})\phantom{\rule{0ex}{0ex}}=p(z-1.5)$

Using z table

$=0.0668\cdot N\phantom{\rule{0ex}{0ex}}=0.0668\cdot 1000$

Probability = 66.8

b) $p(180x300)\phantom{\rule{0ex}{0ex}}=p(180-\frac{240}{40})(x-\frac{\mu}{\sigma})(300-\frac{240}{40})\phantom{\rule{0ex}{0ex}}=p(-\frac{60}{40}z\frac{60}{40})\phantom{\rule{0ex}{0ex}}=p(-1.5z1.5)\phantom{\rule{0ex}{0ex}}=p(z1.5)-p(z1.5)$

Using z table

$=0.9332-0.0668\phantom{\rule{0ex}{0ex}}=0.8664\cdot N\phantom{\rule{0ex}{0ex}}=0.8664\cdot 1000$

Probability = 866.4

c) $p(110x180)\phantom{\rule{0ex}{0ex}}=p((110-\frac{240}{40})(x-\frac{\mu}{\sigma})(180-\frac{240}{40})\phantom{\rule{0ex}{0ex}}=p(-\frac{130}{40}z-\frac{60}{40})\phantom{\rule{0ex}{0ex}}=p(-3.25z-1.5)\phantom{\rule{0ex}{0ex}}=p(z-1.5)-p(z-3.25)$

Using z table

$=0.0668-0.0006\phantom{\rule{0ex}{0ex}}=0.0662\cdot N\phantom{\rule{0ex}{0ex}}=0.0662\cdot 1000$

Probability = 66.2

d) $p(x300)\phantom{\rule{0ex}{0ex}}=1-p(x300)\phantom{\rule{0ex}{0ex}}=1-p(x-\frac{\mu}{\sigma})(300-\frac{240}{40})\phantom{\rule{0ex}{0ex}}=1-p(z\frac{60}{40})\phantom{\rule{0ex}{0ex}}=1-p(z1.5)$

Using z table

$=1-0.9332\phantom{\rule{0ex}{0ex}}=0.0668\ast N\phantom{\rule{0ex}{0ex}}=0.0668\ast 1000$

Probability = 66.8

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