Question

For 12 – 15, round to 2DP.

12. Find the z-score for which 70% of the area is to the right.

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13. Find the z-score that corresponds to P₆

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14. Find the z-scores for which 60% of the distribution’s area lies between − z and z.

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Answer 1

Solution :

12 ) Given that,

Using standard normal table,

P(Z > z) = 70%

1 - P(Z < z) = 0.70

P(Z < z) = 1 - 0.70 = 0.30

P(Z < 0.52 ) =0.30

z = 0.52

13 ) Given that,

Using standard normal table,

P(Z < z) = 6%

P(Z < z) = 0.06

P(Z < 1.55 ) =0.06

z = 1.55

14 ) Given that,

Using standard normal table,

P(-z < Z < z) = 0.60

P(Z < z) - P(Z < z) = 0.60

2P(Z < z) - 1 = 0.60

2P(Z < z) = 1 + 0.60

2P(Z < z) = 1.60

P(Z < z) = 1.60 / 2 = 0.80

P(Z < z) = 0.80

z = 0.84

− z = - 0.84