Answer:
P( Z₁ < z < Z₂) = A(Z₂ ) - A(Z₁) = 0.2182
Step-by-step explanation:
Given Z is a standard normal variable
Given Z₁ = 0.7 and Z₂ = 1.98
P( Z₁ < z < Z₂) = P( 0.7 < z < 1.98)
Here Z₁ and Z₂ both are positive
P( Z₁ < z < Z₂) = A(Z₂ ) - A(Z₁)
= A(1.98) - A(0.7)
= 0.4762 - 0.2580
= 0.2182
Conclusion:-
P( Z₁ < z < Z₂) = A(Z₂ ) - A(Z₁) = 0.2182
The probability that z lies between 0.7 and 1.98 is 0.21
For us to get the probability that z lies between 0.7 and 1.98, we will first need to get the equivalent z-scores for each value.
For the value 0.7, the z-scores is 0.2580
For the value of 1.98, the z-score is 0.4762
As we can see the z-score values are both positive, hence;
P( z₁ < z < z₂) = P(z₂ ) - P(z₁)
P( z₁ < z < z₂) = P(1.98) - P(0.7)
P( z₁ < z < z₂) = 0.4762 - 0.2580
P( z₁ < z < z₂) = 0.2182
Hence the probability that z lies between 0.7 and 1.98 is 0.2182
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