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Q.Using Z-Table, Calculate

a) P(Z<1.20)P(Z < 1.20)
b) P(Z≤1.20)P(Z \le 1.20)
c) P(−0.5≤Z≤1.0)P(-0.5 \le Z \le 1.0)
d) P(−1.0≤Z≤1.0)P(-1.0 \le Z \le 1.0)
Puducherry CbseNCERTSubjective· 2mImportance★★★★★est
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Reading the standard normal table: P(Z<1.20)=P(Z≤1.20)=0.8849P(Z<1.20)=P(Z\le1.20)=0.8849; P(−0.5≤Z≤1.0)=0.5328P(-0.5\le Z\le1.0)=0.5328; P(−1.0≤Z≤1.0)=0.6826.P(-1.0\le Z\le1.0)=0.6826.

Φ(z)=P(Z≤z)\Phi(z)=P(Z\le z) from the standard normal table, and P(a≤Z≤b)=Φ(b)−Φ(a)P(a\le Z\le b)=\Phi(b)-\Phi(a).

Symmetry: Φ(−z)=1−Φ(z)\Phi(-z)=1-\Phi(z). For a continuous variable P(Z<z)=P(Z≤z)P(Z<z)=P(Z\le z).

Steps

  1. (a) From the table, Φ(1.20)=0.8849\Phi(1.20)=0.8849, so P(Z<1.20)=0.8849.P(Z<1.20)=0.8849.

  2. (b) Since ZZ is continuous, P(Z≤1.20)=P(Z<1.20)=0.8849.P(Z\le1.20)=P(Z<1.20)=0.8849.

  3. (c) P(−0.5≤Z≤1.0)=Φ(1.0)−Φ(−0.5).P(-0.5\le Z\le1.0)=\Phi(1.0)-\Phi(-0.5). Here Φ(1.0)=0.8413\Phi(1.0)=0.8413 and Φ(−0.5)=1−Φ(0.5)=1−0.6915=0.3085.\Phi(-0.5)=1-\Phi(0.5)=1-0.6915=0.3085. So =0.8413−0.3085=0.5328.=0.8413-0.3085=0.5328.

  4. (d) P(−1.0≤Z≤1.0)=Φ(1.0)−Φ(−1.0)=0.8413−(1−0.8413)=0.8413−0.1587=0.6826.P(-1.0\le Z\le1.0)=\Phi(1.0)-\Phi(-1.0)=0.8413-(1-0.8413)=0.8413-0.1587=0.6826. …

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