Hey,(adsbygoogle = window.adsbygoogle || []).push({});

I'm new to all this so cut me some slack, but have been trying to work through some questions, and I cant seem to find answers to these questions... Or atleast find the confidence that my answers/working is correct...

1. (Exponential Distribution) Telephone calls arrive at the information desk of a large(What would eb the formula for this one???) I got 0.98347

computer software company at the rate of 15 per hour.

What is the probability that the next call will arrive within 15 minutes?

2. (Normal Distribution) For a group of trucks, it was found on an annual basis that the distance travelled per truck is normally distributed with a mean of 50.0 thousand km and a standard deviation of 12.0 thousand km.

(i) What proportion of trucks can be expected to travel between 34.0 and 38.0

thousand km in the year?

I get around 6.64%, but im not sure that Im doing it right as I tried to do the same thing on the computer and I get something like 6.674438...

(ii) How many km will be travelled by at least 80% of the trucks?

(This one really has me stumpted as im not sure if it means I look at the normal distro curve, and sort of 40% on each side or, do I try and figure out the 20% value there, and say that at least 80% of the trucks travel >than that distance. The latter makes more sense to me...) I tried the latter and I got 39,920km

3. (Binomial Distribution) A student is taking a multiple-choice exam in which each

question has four choices.

Assuming that he/she has no knowledge of the correct answers to any of the questions, he decided on a strategy in which he will place four balls (marked A, B, C and D) into a box.

He randomly selects one ball (with replacement)for each question. The marking on the ball will determine hisanswer to the question.

There are five multiple-choice questions on the exam. What is the probability the he will get:

(i) five questions correct?

(ii) at least four questions correct?

I think i have this one sorted now, i)=1/1024 ii)=5/1024

4.(Poisson Distribution) Assume that the number of network errors experienced in a day on

a local area network (LAN) is distributed as a Poisson random variable. The average

number of network errors experienced in a day is 2.4. What is the probability that in any

given day

(i) exactly one network error will occur?

(ii) two or more network errors will occur?

i) This one I get ~21.7%

ii) I did 1-P(X=2) = ~78%

Any help please?

Thanks,

NW

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# Advice on Exponential, Binomial, & Normal Distributions

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