Question

A statistical analysis of​ 1,000 long-distance telephone calls made by a company indicates that the length of these calls is normally​ distributed, with a mean of 240 seconds and a standard deviation of 30 seconds. Complete parts​ (a) through​ (d).

a. What is the probability that a call lasted less than 180 ​seconds? The probability that a call lasted less than 180 seconds is nothing. ​(Round to four decimal places as​ needed.)

b. What is the probability that a call lasted between 180 and 290 ​seconds? The probability that a call lasted between 180 and 290 seconds is nothing. ​(Round to four decimal places as​ needed.)

c. What is the probability that a call lasted more than 290 ​seconds? The probability that a call lasted more than 290 seconds nothing. ​(Round to four decimal places as​ needed.)

d. What is the length of a call if only 0.5 % of all calls are​ shorter? 0.5​% of the calls are shorter than nothing seconds. ​(Round to two decimal places as​ needed.)

Answer 1

Solution :

Given that ,

mean = = 240

standard deviation = = 30

a)P(x < 180) = P[(x - ) / < (180- 240) / 30 ]

= P(z < -2)

= 0.0228

The probability that a call lasted less than 180 seconds is 0.0228

b) P(180 < x < 290) = P[(180 - 240)/30 ) < (x - ) /  < (290 - 240) / 30) ]

= P(-2 < z < 1.67)

= P(z < 1.67) - P(z < -2)

= 0.9525 - 0.0228

0.9297

The probability that a call lasted between 180 and 290 seconds is 0.9297

c) P(x > 290) = 1 - P(x < 290)

= 1 - P[(x - ) / < (290 - 240) /30 )

= 1 - P(z < 1.67)

= 1 - 0.9525

0.0475

The probability that a call lasted more than 290 seconds is 0.0475

d) Using standard normal table ,

P(Z < z) = 0.5%

P(Z < z) = 0.005

P(Z < 2.58) = 0.005

z = 2.58

Using z-score formula,

x = z * +

x = 2.58 * 30 + 240 = 317.40

0.5​% of the calls are shorter than nothing seconds is 317.40