Math
posted by Nora 512 .
Can someone answer a question similar to a and b so I can use as an example.
Consider the following binomial random variables.
(a) The number of tails seen in 47 tosses of a quarter.
(i) Find the mean. (Give your answer correct to one decimal place.)
(ii) Find the standard deviation. (Give your answer correct to two decimal places.)
(b) The number of lefthanded students in a classroom of 59 students (assume that 17% of the population is lefthanded).
(i) Find the mean. (Give your answer correct to one decimal place.)
(ii) Find the standard deviation. (Give your answer correct to two decimal places.)

I have 24 questions like the two above and if anyone can help I will use them as examples and work the other ones. I do so much better with examples. Thank you

For a binomial distribution with N bernoulli trials with probability of success p (i.e. failure q = 1p),
the following properties can be proved:
mean = Np
variance = Npq
Take the squareroot of variance to get standard deviation.
In 15 tosses of a fair dime, N=15, p=0.5, q=10.5=0.5
so mean = 15*0.5=7.5
standard deviation = sqrt(Npq)=√(15*.5*.5)=1.936 
Thank you, I have worked 3 out like the first one, but how do you do one with % just show me example as the one above, don't have to use those numbers just something I can go by...I did get the other 3 I worked out right..Thanks again

17% are lefthanded means
P(lefthanded)=0.17
(remember 17%=17/100=0.17)
so
N=59, p=0.17, q=10.17=0.83
you can take it from here! 
Ok I have got this far and missed the last two on standard deviation, can you look at these and tell me how I missed them?
(23) The number of cars found to have unsafe tires among the 379 cars stopped at a roadblock for inspection (assume that 15% of all cars have one or more unsafe tires).
(i) Find the mean. (Give your answer correct to one decimal place.)
Correct: Your answer is correct. .
56.9 by 379 x .15 =
(ii) Find the standard deviation. (Give your answer correct to two decimal places.)
Incorrect: Your answer is incorrect. . answer 6.79 by sqrt(379 x .15x.81)
(24) The number of melon seeds that germinate when a package of 60 seeds is planted (the package states that the probability of germination is 0.89.
(i) Find the mean. (Give your answer correct to one decimal place.)
Correct: Your answer is correct. 53.40 by 60 x 0.89 =
(ii) Find the standard deviation. (Give your answer correct to two decimal places.)
Incorrect: Your answer is incorrect. sqrt(60 x .89x.18) = 3.07 and tried again 3.10 
For 23)(ii) I have
σ=√(npq)=√(379*0.15*0.85)=6.95
For 24(ii)
p=0.89,
q=10.89=0.11
σ=√(npq) 
Ok (23) I got the standard deviation as 1.99 and on (24) I got standard deviation as 0.73. I have worked and worked these and I am missing something if this is not right.
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