Question

3. A company manufactures square boards for the DIY trade. The length of side of each board is normally distributed with mean 1.5m and standard deviation 5mm, independently of all other boards. What is the probability that the area of a randomly chosen board will be less than 2.24m2? Find, approximately, the probability that among 60 such boards, more than 10% will have areas under 2.21m2

Answer 1

Answer) Here, mean = 1.5 m and standard deviation = 5 mm

Now the area should be less than 2.24 m^2

as the area of the square = square of the length

Therefore the length of a square with area 2.24 m^2 = under root of 2.24

which is equivalent to 1.49

so we need to find p(x<1.49)

here we need to find z score first and z = (x-mean)/standard deviation

mean = 1.5 and standard deviation = 0.005 (as 5mm = 0.005 m)

Therefore z = (1.49-1.5)/0.005

= -0.67

therefore, p(x<1.49) = p(z<-0.67)

= 0.2514 (from z table -0.67 = 0.2514)

Therefore, p(x<1.49) or the probability that the area of randomly chosen board will be less than 2.24 m^2 is 0.2514

B)

Answer)

As there are fixed number of trials and probability of each and every trial is same and independent of each other

Here we need to use the binomial formula

P(r) = ncr*(p^r)*(1-p)^n-r

Ncr = n!/(r!*(n-r)!)

N! = N*n-1*n-2*n-3*n-4*n-5........till 1

For example 5! = 5*4*3*2*1

Special case is 0! = 1

P = probability of single trial = 0.2514

N = number of trials = 60

R = desired success = more than 10% = more than 6

We know that sum of all the probabilities is = 1

So, P(more than 6) = 1 - (P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6))

= 0.99706742383