If z is a standard normal variable, find the probability that z lies between 0.7 and 1.98.

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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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

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step-by-step explanation:

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7 hours, 0 minutes, 0 seconds

or:  

0.04

total weeks

or:  

0.29

total days

or:  

7.00

total hours

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420.00

total minutes

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25,200.00

total seconds

step-by-step explanation: