Question

The Colorado Mining and Mineral Company has 1000 employees engaged in its mining operations. It has been estimated that the probability of a worker meeting with an accident during a 1-yr period is 0.08. What is the probability that more than 70 workers will meet with an accident during the 1-yr period?

a. 0.8665
b. 0.9089
c. 0.3585
d. 0.3827
e. 0.6548

Answer 1

a. 0.8665

Explanation:

Given:

probability 'p'=0.08

no. of employees 'n'=1000

The expected value of the binomial distribution is determined by:

E(x)= μ= np

μ= 1000 x 0.08 = 80

standard deviation 'σ' is given by:

σ = √np(1-p) => √1000 x 0.08 x 0.92

σ=8.58

By using continuity correction, we have P(X≥70+0.5) = P(X≥70.5) that is 1 subtracted by the pvalue of Z when X = 70.5.

Therefore,

Z= X-μ/σ =>(70.5-80)/8.85

Z=-1.11

the p-value of Z=-1.11 is 0.1335

S0, 1 - 0.1335 = 0.8665

the probability that more than 70 workers will meet with an accident during the 1-yr period is 0.8665