Question

statistics question:
john is going on a 5-day holiday to Costa Packet. the travel brochure says that on average 5 out of every 7 days are sunny at costa packet. each days weather is independent of the weather on all preceding days.

a)there are two values, n and p, that you need to use in binomial distribution, write down the value of n and the value of p.

b) calculate the probability that john has 2, or less, sunny days on his holiday?
you may use:
(p+q)⁵ = p⁵ + 5p⁴q + 10p³q² + 10p²q³ + 5pq⁴ + q⁵

c)what is the most likely number of sunny days that john will get on his holiday?

Answer 1

well in the front of your textbook you should have a z-score chart and that is exactly what this is is a z-score question
first of all z-score counts how many standard deviations a number is away from the mean..
so what you need to do is take the mean-(number given) / standard dev

this will give you the z-score and from there you look that number up on the chart and it will tell you the percentage (Going down..so if it says higher/more/greater than you need to subtract the number from 100)..

firstly
a) less than 600$
z-score = 680-600 / 138 = -.5797 standard devs away from the mean..
that zscore gives you a percentage of 30.5%
b)more than 700$
z-score= 680-700/138 = .1449 standard devs away from the mean.. that zscore gives you a percentage of 92.65%
c)between 600$ and 700$
for these you take the lower zscore percentage and subtract it from the higher zscore percentage.
600$ = 30.5%
700$ = 92.65%
92.65%-30.5% = 62.15%

those are your three answers hope i cleared things up a little :)
you can always use an online zscore chart.
Source(s):
http://www.epatric.com/documentation/sta …

Answer 2

but i got n=7 and p=5/7 :/

Answer 3

u r correct

Answer 4

me?

Answer 5

ya!

Answer 6

would youu be able to help me with the rest please? :)

Answer 7

once youve answered it

Answer 8

what on earth is going on here?

Answer 9

a lot of words again that have nothing to do with the question. this character needs to be stopped

Answer 10

first off you are right about p since it says sunny five out of seven days \(p=\frac{5}{7}\)
but you are mistaken about n
n is the number of days on the trip (the number of elements in the "sample space") and it is 5

Answer 11

@satellite73 , couldn't agree with you more !

Answer 12

so we have \[n=5,p=\frac{5}{7}, 1-p=\frac{2}{7}\]

Answer 13

it is one thing to give a wrong answer because you are mistaken
i have certainly done that. it is quite another to copy and paste some non related garbage and post it as an answer, then say something random

Answer 14

so now will it follow binomial distribution?

Answer 15

@satellite73

Answer 16

@islacameron if you are still there, we can complete this problem if you like. once we have
\[n=5,p=\frac{5}{7}, q=1-p=\frac{2}{7}\]

Answer 17

two or less sunny days means
\[P(x=0)+P(X=1)+P(X+2)\]

Answer 18

you have
\[(\frac{2}{7})^5+6(\frac{2}{7})^4\times \frac{5}{7}+10(\frac{2}{7})^3\times (\frac{5}{7})^2\]

Answer 19

normally you would have to compute the binomial coefficents but they were given to you in this line above
\[(p+q)⁵ = p⁵ + 5p⁴q + 10p³q² + 10p²q³ + 5pq⁴ + q⁵ \]

Answer 20

Easiest of all
c)what is the most likely number of sunny days that john will get on his holiday?

Answer -->5

Answer 21

@islacameron if you have any questions let me know, i will be happy to answer about this problem. ignore @Rohangrr posts, they are almost always random

Answer 22

i htink most likely sunny days needs to be computed.
sunny all five days has probability \((\frac{5}{7})^5=.186\) rounded, whereas 4 sunny days is \(5\times (\frac{5}{7})^4\times \frac{2}{7}=.372\) rounded

Answer 23

evidently the probability of 4 is larger than the probability of all five
we can also compute the probability of exactly 3 sunny days via \(10\times (\frac{5}{7})^3(\frac{2}{7})^2=.297\) rounded

Answer 24

unless you were talking about someone that gave the answer then in xthat case im am terribly sorry and take all of that back! :)

Answer 25

if you are addressing me, my comments about random answers were directed to rohangrr's post, not your question
i hope i answered your question completely, and if you have any question about what i wrote i would be happy to explain in detail

Answer 26

ahh well yeah in that case i apologise and that you a lot for your help and answer, sorry

Answer 27

that nonsense starting with "z scores" has nothing to do with the problem

Answer 28

did you get that \(n=5, p=\frac{5}{7}\) ?

Answer 29

yeahh, thanks

Answer 30

and the rest as well for the binomial distribution?

Answer 31

yeah thanks :)

Answer 32

ok, good.