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dimulka [17.4K]
4 years ago
14

Find a value of the standard normal random variable Z, Call it Z0, such that

Mathematics
1 answer:
katrin [286]4 years ago
5 0
For the answer to this questions,<span>a. P (z ≤ z0) = 0.0401=P(z ≥ z0) = 1-0.0401 = 0.9599 = P(z ≤ -z0) = 0.9599 
From tables z0 = -1.75 

b. P (-z0 ≤ z ≤ z0) = .95 = P (z ≤ z0)- P (z ≤ -z0) = P (z ≤ z0)- P (z ≥ z0) = 
P (z ≤ z0)-(1- P (z ≤ z0)) 
P (z ≤ z0) = (0.95+1)/2=0.975 
From tables z0 = 1.96 

c. P (-z0 ≤ z ≤ z0) = 0.90 
the procedure is the same that exercise b P (z ≤ z0) = (0.9+1)/2=0.95 
From tables the nearest value is z0 = 1.64 

</span>d. P (-z0 ≤ z ≤ 0) = 0.2967= P (z ≤ 0) - P (z ≤ -z0) = P (z ≤ 0) - P (z ≥ z0) = 
<span>P (z ≤ 0) - (1- P (z ≤ z0)) </span>
<span>P (z ≤ z0) = 0.2967 + 1 - P (z ≤ 0)= 0.2967 + 1 - 0.5 = 0.7967 </span>
<span>From tables z0 = 0.83 
</span><span>
I hope my answer helped you

</span>
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