Question

30. The weight of Canada Post letters is normally distributed with a mean of 2 oz. and a standard deviation of 0.5 oz Use this information for this question and the next one. If I select at random two Canada Post letters, what is the probability that they will both weigh less than 1 oz.? (a) 0.0005 (b) 0.0228 c) 0.0456 d 0.4772 (e) 0.8413 Solution: (a) The probability that one letter weigh less than 1 oz. is P(X < 1)-P(Z < 20.0228. The probability that they will both weigh less than 1 oz is 0.02282 (multiplication rule for independent events)
For Q31, we can see the SD is root 22.280, but where is 22.280 come from?

Answer 1

In normal approximation,

mean = n*p

Variance = n*p*q

Std deviation = sqrt(n*p*q)

Where n = 1000, p = P(z<1) = 0.0228 , q = 1- p = 0.9772

Substituting the values of n, p, q

Mean = 1000*0.0228= 22.8,

Variance = 1000* 0.0228*0.9772 = 22.28

Std deviation = sqrt(1000*0.0228*0.9772) = sqrt(22.28) = 4.72

This is how you will get the standard deviation.