Answer:
0.0202
Explanation:
μ=15, σ=0.8; n=30
p(x≤ 14.7) = p(Ζ ≤ 14.7-15/0.8÷√30)
p(x≤ 14.7) = p(Ζ ≤ -2.05)
p(x≤ 14.7) = 0.0202
So, the company's claim is not reasonable because the probability that a sample of 30 batteries would have a lifetime of almost 14.7 hours is about 0.0202.