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Statistics...help!

I have exam tomorrow for paper 6 statistics, and here are the questions that I am still stuck and couldn't do them..can someone please help me! Thanks!

3)Sam caught three different types of fish and recorded the type and mass (correct to the nearest 0.01 kg) of each fish caught. At 4 p.m. he summarised the results as follows.
Type of fish Perch Tench Roach
Number of fish 1 0 1
Mean mass(kg) = 1
Standard deviation (kg) =0
Before leaving the waterside, Sam catches one more fish and weighs it. He then announces that if this extra fish is included with the other two fish he caught, the standard deviation is 1.00 kg. Find the mass of this extra fish.

----Let x be the mass of the third fish's mass.
Sqrt (2+x^2)/3 - [(2+x)/3]^3 = 1 and then I tried to find the value of x, but I failed to do so.

4)Show that, when two fair dice are thrown, the probability of obtaining a 'double' is 1/6, where a 'double' is defined as the same score on both dice. (done) Four players play a board game which requires them to take it in turns to throw two fair dice. Each player throws the two dice once in each round. When a 'double' is thrown the player moves forward six squares. Otherwise the player moves forward one square. Find
c)the probability that a 'double' occurs exactly once in 4 of the first 5 rounds.

----For question 4d, I use C(4, 1) * (1/6) * (5/6)^3 * (5/6)^5 but it gives me the wrong answer.

5)
a)A fair coin is tossed 8 times. Calculate the probability that the first 4 tosses and the last 4 tosses result in the same number of heads. (ans. =0.273)
b)Two teams each consist of 3 players. Each player is in a team tosses a fair coin once and the team's score is the total number of heads thrown. Find the probability that the teams have the same score. (ans. =0.313)

----For question five a, I used C(4, r) * 0.5^r * 0.5^4-r where r = 0, 1, 2, 3, 4 for the first four tosses but multiply by two for the last fur tosses but it gives me the wrong answer as well.
For question five b, I used C(3, r) * 0.5^r * 0.5^3-r where r=0,1,2,3) and then times two as for the team B as well, but again it's wrong again.

6)Six hens are observed over a period of 20 days and the number of eggs laid each day is summarised in the following table.
Number of eggs 3 4 5 6
Number of days 2 2 10 6
State the probability that a randomly chosen hen lays an egg on a given day. (ans. =5/6)
Calculate the expected frequencies of 3, 4, 5, and 6 eggs. (ans. = 1.07, 4.02, 8.04, 6.70)

-----For question six, what is the table for? Is it useful to calculate the probability and how do I calculate the expected frequency?

7)In an experiment, a fair cubical dice was rolled repeatedly until a six resulted, and the number of rolls was recorded. The experiment was conducted 60 times.
b)Find the expected frequency for two rolls correct to one decimal place. (ans. =8.3)

----For question 7, it asks for the expected frequency for two rolls. But the two rolls can be either (1, 2), ( 1, 3)......(6, 1)....and so on, wouldn't it be quite long? Sorry but I don't know how to do it. I was just giving my thoughts.

8)The score S on a spinner is a random variable with distribution given by P(S=s)=k (s=1,2,3,4,5,6,7,8), where k is a constant. Find the value of k.

-----I did it but I get the wrong answer.
k+2k+3k+4k+5k+6k+7k+8k = 1
36k = 1
k =1/36 (But the answer given is 1/8)
Reply 1
Is the first one 720? (If it is method is (5!/2!)*(4!/2!), as the letters are repeated 2 N's nd 2 a's). Havn't got time to look at others,
Jack
Reply 2
authecroix123

9)A cubical dice is biased so that the probabilty of any particular score between 1 and 6(inclusive) being obtained is proportional to that score. Find the probability of scoring a 1.


P( Score =1) =k
P(Score =2) =2k
P(Score =3)= 3k
etc so
k +2k +3k+4k+5k+6k=1
21k=1
k=1/21
Probability of scoring one = 1/21

For question 10
P( Score =1) =k
P(Score =2) =k/2
P(Score =3)= k/3
so k +k/2+k/3+k/4+k/5+k/6=1
Multy by 60
60k+30k+20k+15k+12k+10k=60
90k+ 35k + 22k=60
(125+22)k=60
k=60/147 Probability of scoring one = 60/147
Reply 3
Yes thank you so much for Jack, Carol05 and m277 for helping me so much!
Reply 4
(1)
The consonants are G, N, N, R, T, which can be arranged in 5!/2! = 60 ways.
The vowels are A, A, E, I, which can be arranged in 4!/2! = 12 ways.

Answer = 60*12 = 720

(2)
The consonants are B, M, S, S, T, which can be arranged in 5!/2! = 60 ways.
The vowels are A, E, I, O, U, which can be arranged in 5! = 120 ways

We can arrange the letters as (i) VCVCVCVCVC or (ii) CVCVCVCVCV ("V" = vowel, "C" = consonant). There are 60*120 = 7200 arrangements for each of these two possibilities.

Answer = 2*7200 = 14400
Reply 5
Thanks Jonny! :smile: Just have 7 hours more before I go to school and have my physics paper 1 then maths paper 6!