# Find the probabilities for each, using the standard normal distribution.

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Question : Find the probabilities for each, using the standard normal distribution.
1. P(0<z<1.96)
P(low z-score < z < high z-score) = P(hig z-score) - P(low z-score)
From the Z distribution table we get
Probability for z =1.96 0.975
& Probability for z =0 0.5
Therefore , P(0<z<1.96) = P(z=1.96)-P(z=0)
= 0.975 - 0.5 = 0.475
2. P(-2.46<z<1.74)
P(low z-score < z < high z-score) = P(hig z-score) - P(low z-score)
From the Z distribution table we get
Probability for z scores as follows
P( z =-2.46) = 0.00695 & P( z =1.74) = 0.95907
Therefore, P(-2.46<z<1.74) = P(z=1.74)-P(z=-2.46)
= 0.95907 - 0.00695 = 0.95212
3. P(1.46 < z < 2.97)
P(low z-score < z < high z-score) = P(hig z-score) - P(low z-score)
From the Z distribution table we get
Probability for z scores as follows
P( z =1.46) = 0.92785 & P( z =2.97) = 0.99851
Therefore, P(1.46 < z < 2.97) = P(z=2.97) - P(z=1.46)
= 0.99851 - 0.92785 = 0.07066
Hence P(1.46 < z < 2.97) = 0.07066 Answer

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