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You may need to use the appropriate appendix table to answer this question. Given that z is a standard normal random variable

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Answer #1

Solution :

Given that,  

Using standard normal table ,

(a)

P(-1.98 \leq z \leq 0.47)

= P(z \leq 0.47) - P(z \leq -1.98)

= 0.6808 - 0.0239

= 0.6569

P(-1.98 \leq z \leq 0.47) = 0.6569

(b)

P(0.54 \leq z \leq 1.24)

= P(z \leq 1.24) - P(z \leq 0.54)

= 0.8925 - 0.7054

= 0.1871

P(0.54 \leq z \leq 1.24) = 0.1871

(c)

P(-1.75 \leq z \leq -1.02)

= P(z \leq -1.02) - P(z \leq -1.75)

= 0.1539 - 0.0401

= 0.1138

P(-1.75 \leq z \leq -1.02) = 0.1138

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