The marks in an examination for a physics paper have normal distribution with mean μ and variance σ2 . 10% of the students obtain more than 75 marks and 20% of them obtained less than 40 marks. find the value of μ and σ

The mean of the distribution is about 53.9 and standard deviation is about 16.5 marks.

Step-by-step explanation:

Use the trick of transforming the given random variable one that is standard normal distributed, aka a z-score, then look up the two percentile values in z-score tables to get two equations with two unknowns.

So, let x be the variable describing the marks of a student, and

the standardized equivalent of that (mu - mean, sigma - standard deviation).

We are looking for values of mu and sigma. At this point we'd be out of luck, but, wait, we're given two bits of info: the 10% point (aka, 90th percentile) and the 20% point (can be interpreted as 100-20th percentile). For each point we can use z-score tables to look up the corresponding values of z (just search for z tables). I found:

z-score for the 10% point: z_10=1.28

z-score for the 20% point: z_20=-0.84

That gives us two equations:

and can be solved for mu and sigma (do the work on your end, I am showing my result):

The mean of the distribution is about 53.9 and standard deviation is about 16.5 marks.

fa tha first one you gonna add y to both sides

it look like this

x-y+y=6+y

x=y+6

therefore your answer is

x =y+6

fa tha second one

you gonna flip tha equation

it gonna look like x-4y=y

then add 4 to both sides

it gonna look like tha

x-4 +4 =y +4

so therefore tha answer is x=y+4

step-by-step explanation:

1,407.5 hours

step-by-step explanation:

we first convert all the time to hours and then sum them up.

there are 24 hours in 1 day, in 58 days, there are 24 x 58 = 1,392 hours.

there are 60 minutes in 1 hour, thus, 30 minutes will be equivalent to 30 / 60 = 0.5 hours.

therefore, the number of hours it will take mercury to complete a full rotation is 1,392 + 15 + 0.5 = 1,407.5 hours