There are two questions within this post one labeled #15 and the other one #16 1

Question

There are two questions within this post one labeled #15 and the other one #16
15) Which of the following ranges of z values corresponds to a probability of 0.8213?
The answer is -1.13?z?1.65
but I'm kind of confused on how to solve it. I know I have to look up the z values from a chart but I'm still confused

16) One model of an imported automobile is known to have defective seat belts. An extensive study has found that 17% of the seat belts were incorrectly installed at the factory. A car dealer has just recieved an shipment of 196 cars of this model. What is the probability that in this sample:
(a) 22 or fewer have defective seat belts?
(b) 30 or more have defective seat belts?
(c) between 25 and 50 have defective seat belts?
*Answers are
(a) P(r?22)=P(x?22.5)=P(z?-2.06)=0.0197
(b)P(r?30)=P(x?29.5)=P(z?-0.73)=0.7673
(c)P(25?r?50)=P(24.5?x?50.5)=P(-1.68?z?3.27)=0.9530

Explanation / Answer

(15)

From Z-distribution table :

Z=-1(0.8213) = 0.920

From the option , the choice is :

-1.13 < z < 1.65

(16)

number of defective = P = 0.17

= p*(1-p)/196

sample:
(a) 22 or fewer have defective seat belts?

p1=22/196

P(r<22)=P(X<p1) =P(X-p/<p1-p/)

=P(z<-2.06)=0.0197

(b)30 or more have defective seat belts?

p1=30/196

P(r>30)=P(X>p1) =P(X-p/>p1-p/)

=P(z>-0.73)=0.7673

(c) between 25 and 50 have defective seat belts?

p1=30/196

p2 = 50/196

P(25<r<50)=P(p1<X<p2)

=P(p1-p/<z<p2-p/)

=0.9530

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