If X is a normal variate with mean 30 and standard deviation 5- what is Probability 26- X- 34- Given A z - 0-8 - 0-2881- where A represents area-

We know that, nbsp; nbsp; nbsp;Z=X μσ nbsp; nbsp; nbsp;Z_1=26 305= 0.8 and nbsp; nbsp; nbsp;Z_2=34 305=0.8 ∴ P(26≤ X≤ 34)=P( 0.8≤ ...

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Normal Distribution Properties

If X is a nor...

Question

If X is a normal variate with mean 30 and standard deviation 5, what is Probability $(26 \leq X \leq 34),$ . Given A (z = 0.8) = 0.2881, where A represents area.

0.2881

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0.5762

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0.8181

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0.1616

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Solution

The correct option is B 0.5762
We know that, $Z = \frac{X - μ}{σ}$

$Z_{1} = \frac{26 - 30}{5} = - 0.8$

and $Z_{2} = \frac{34 - 30}{5} = 0.8$

$∴ P (26 \leq X \leq 34) = P (- 0.8 \leq Z \leq 0.8)$

$= 2 P (0 \leq Z \leq 0.8)$
$= 2 A r e a (0.8)$
= 2 × 0.2881
= 0.5762

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If X is a normal variate with mean 30 and standard deviation 5, what is Probability (26≤X≤34),. Given A (z = 0.8) = 0.2881, where A represents area.

The correct option is B 0.5762We know that, Z=X−μσ Z1=26−305=−0.8 and Z2=34−305=0.8 ∴ P(26≤X≤34)=P(−0.8≤Z≤0.8) =2P(0≤Z≤0.8) =2Area(0.8) = 2 × 0.2881 = 0.5762

If X is a normal variate with mean 30 and standard deviation 5, what is Probability $(26 \leq X \leq 34),$ . Given A (z = 0.8) = 0.2881, where A represents area.

The correct option is B 0.5762
We know that, $Z = \frac{X - μ}{σ}$

$Z_{1} = \frac{26 - 30}{5} = - 0.8$

and $Z_{2} = \frac{34 - 30}{5} = 0.8$

$∴ P (26 \leq X \leq 34) = P (- 0.8 \leq Z \leq 0.8)$

$= 2 P (0 \leq Z \leq 0.8)$
$= 2 A r e a (0.8)$
= 2 × 0.2881
= 0.5762