Question

A company produces precision 1-meter (1000 mm) rulers. The actual distribution of lengths of rulers produced by this company is approximately normal with mean 1000 and standard deviation 0.02mm.

(a) Approximately 16% of the rulers are less than what length?  

(b) Suppose you take a random sample of 5 rulers. What is the probability that the average of the sample is greater than 1000.01 mm?  

Answer 1

Solution :

(a)

Using standard normal table ,

P(Z < z) = 16%

P(Z < -0.99) = 0.16

z = -0.99

Using z-score formula,

x = z * +

x = -0.99 * 0.02 + 1000 = 999.98

999.98 length

(b)

= / n = 0.02 / 5 = 0.0089

P( > 1000.01) = 1 - P( < 1000.01)

= 1 - P[( - ) / < (1000.01 - 1000) / 0.0089]

= 1 - P(z < 1.1236)

= 1 - 0.8694

= 0.1306

Probability = 0.1306