My Math Forum Truck Drivers and Statistic Question?

 April 3rd, 2012, 04:54 PM #1 Newbie   Joined: Jan 2012 Posts: 6 Thanks: 0 Truck Drivers and Statistic Question? How do I solve this problem? And this one, a statement has to be written along with it as well. According to a study of 90 truckers, a trucker drives, on average, 540 miles per day. If the standard deviation of the miles driven per day for the population of truckers is 40, find the 99% confidence interval of the mean number of miles driven per day by all truckers?
May 4th, 2012, 03:45 AM   #2
Newbie

Joined: May 2012

Posts: 13
Thanks: 0

Re: Truck Drivers and Statistic Question?

Quote:
 Originally Posted by gigsaw According to a study of 90 truckers, a trucker drives, on average, 540 miles per day. If the standard deviation of the miles driven per day for the population of truckers is 40, find the 99% confidence interval of the mean number of miles driven per day by all truckers?
From the problem we have:

$n= 90;\; \bar {x} = 540 miles/day ;\; s = 40 miles/day$

and we need to calculate 99 % CI of the $\mu$

we also know that:

$Z= \frac {\bar {X} - \mu} {\frac {s} {\sqr {n}}}$ and $P(-2.576\le {Z} \le 2.576)= P(-2.576\le {\frac {\bar {X} - \mu} {\frac {s} {\sqr {n}}}} \le 2.576)=0.99$ then:

99 % CI of $\mu= (540 - 2.576 \frac {40}{\sqr {90}}\; ;\; 540 + 2.576 \frac {40}{\sqr {90}})= P(529.139 \; ; \;550.861)$ miles/day

 January 7th, 2014, 11:48 PM #3 Newbie   Joined: Jan 2014 Posts: 2 Thanks: 0 Re: Truck Drivers and Statistic Question? Some questions strikes my mind too. Is it important to get 4x4 education from any driving school? What is its benefit? Anyone can help?
 April 10th, 2014, 01:57 AM #4 Newbie   Joined: Apr 2014 From: USA Posts: 1 Thanks: 0 Can you please tell me which is best Light Vehicle Driver Training school available in US. If anyone have idea? I need to take the classes so that i can start my own business soon.
April 10th, 2014, 04:09 PM   #5
Newbie

Joined: Jan 2014

Posts: 23
Thanks: 0

Quote:
 $Z= \frac {\bar {X} - \mu} {\frac {s} {\sqr {n}}}$ and $P(-2.576\le {Z} \le 2.576)= P(-2.576\le {\frac {\bar {X} - \mu} {\frac {s} {\sqr {n}}}} \le 2.576)=0.99$ then: 99 % CI of $\mu= (540 - 2.576 \frac {40}{\sqr {90}}\; ;\; 540 + 2.576 \frac {40}{\sqr {90}})= P(529.139 \; ; \;550.861)$ miles/day
It requires the assumption that $\displaystyle X$ is normally distributed doesn't it? (just curious )

 Tags drivers, question, statistic, truck

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# according to a study of 90 truckers, a trucker drives, on average, 540 miles per day. if the standard deviation of the miles driven per day for the population of truckers is 40, find the 99% confidence interval of the mean number of miles driven per day

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post green21317 Algebra 2 March 17th, 2014 08:45 AM itamaratento Applied Math 2 January 30th, 2014 10:02 AM kingkos Advanced Statistics 1 October 2nd, 2012 12:08 PM melisavictor Algebra 2 February 27th, 2012 11:58 PM EatAnApple Advanced Statistics 1 September 11th, 2010 03:29 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top