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 left-handed students in a classroom of 59 students (assume that 17% of the population is left-handed).
(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 = 1-p),

the following properties can be proved:
mean = Np
variance = Npq
Take the square-root of variance to get standard deviation.

In 15 tosses of a fair dime, N=15, p=0.5, q=1-0.5=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 left-handed means

P(left-handed)=0.17
(remember 17%=17/100=0.17)
so
N=59, p=0.17, q=1-0.17=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=1-0.89=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.