The normal distribution Mixed exercise 3

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1 The normal distribution Mixed exercise 3 ~ N(78, 4 ) a Using the normal CD function, P( 85) (4 d.p.) b Using the normal CD function, P( 8) The probability that three men, selected at random, all satisfy this criterion is 3 P( 8) (4 d.p.). c Using the inverse normal function, P( h).5 h To the nearest centimetre, the height of a door frame needs to be at least 88 cm. W ~ N(3.5,. ) a Using the normal CD function, P( W 3) The percentage of sheets weighing less than 3kg is.8% (3 s.f.). b Using the normal CD function, P(3.6 W 34.8) So 5.% of sheets satisfy Bob s requirements. 3 T ~ N(48, 8 ) a Using the normal CD function, P( T 6) The probability that a battery will last for more than 6 hours is.668 (4 d.p.). b Using the normal CD function, P( T 35) The probability that a battery will last for less than 35 hours is.5 (4 d.p.). c Use the binomial distribution X B(3,.58...) Using the binomial CD function, P( X 3) The probability that three or fewer last less than 35 hours is.935 (4 d.p.). Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

2 4 X ~ N(4, ) a 3 P( X 3).5 P Z.5 Using the inverse normal function, z so (3 s.f.) b Using the normal CD function, P( X ) (3 d.p.) c d P( X d). P Z. Using the inverse normal function, z d 4 so d (3 s.f.) Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

3 5 L ~ N(, ) a 4 P( L 4). P Z. Using the inverse normal function, z so So the standard deviation of the volume dispensed is 8.6 ml (3 s.f.). b Using the normal CD function, P( L ) The probability that the machine dispenses less than ml is. (3 s.f.). c c P( L c). P Z. Using the inverse normal function, z c so c To the nearest millilitre, the largest volume leading to a complaint is 9 ml. Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 3

4 6 a P( X ).5 and P( X 4).75 Using the inverse normal function (or the percentage points table), P( X ).5 P Z.5 z P( X 4).75 P Z.75 z So.6745 ( ) and ( ) ( ) ( ): Substituting into ( ): So 3 and 4.8 (3 s.f.) b Using the inverse normal CD function with 3 and , P( X a). a and P( X b).9 b So the % to 9% interpercentile range is P( 5). P 5 Z. z P( 4).5 P Z.5 z So Subtract cm (3 s. f.) cm 8 a b T ~ N(8, ) Using the normal CD function, P( T 85) (4 d.p.) S ~ N(, 5 ) Using the normal CD function, P( S 5) (4 d.p.) c The student s score on the first test was better, since fewer of the students got this score or higher. Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 4

5 9 J ~ N(8, ) a P( J ).3 P Z.3 Using the inverse normal function, z so g 3 s.f.. The standard deviation is 4.5 g (3 s.f.). b Using the normal CD function, P( J 5) (3 d.p.) c Use the binomial distribution X B(5,.5) Using the binomial CD function, P( X ) (4 d.p.) T N(, 3.8 ) and P( T 5).446 X a P( T 5).446 P Z. 446 z. 7 5 so minutes (3 s.f.) b P( T 5) P Z 3.8 P( Z.93...) (4 d.p.) Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 5

6 T N(, ) Using the inverse normal function, 7 P( T 7).986 P Z.986 z P( T 5.). P Z. z So. 7 ( ) and ( ) ( ) ( ): Substituting into ( ): So the mean thickness of the shelving is 6. mm and the standard deviation is.398 mm (3 s.f.). Let X = number of heads in 6 tosses of a fair coin, so X ~ B(6,.5). Since p =.5 and 6 is large, X can be approximated by the normal distribution where and So Y N(3,5) P( X 5) P ( Y 4.5) (3 s.f. ) Y N(, ), 3 a The distribution is binomial, B(,.4). The binomial distribution can be approximated by the normal distribution when n is large (> 5) and p is close to.5. ere n = and p =.4 so both of these conditions are satisfied. b np.4 4 and np( p ) (4 s.f.) c P( X 5) P( Y 49.5) (3 s. f.) 4 a P( X ).467 (4 s.f.) or.47 (4 d.p.) b The distribution satisfies the two necessary conditions for an approximation by a normal distribution to be valid: n = is large (> 5) and p =.46 is close to.5. np and np( p ) (4 s.f.) Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 6

7 4 c Y N(55., 5.46 ) Using the normal CD function, P( X 65) P(64.5 Y 65.5) Percentage error.7% a The distribution satisfies the two necessary conditions for an approximation by a normal distribution to be valid: n = 3 is large (> 5) and p =.6 is close to.5. b np and np( p ) (4 s.f.) So Y N(8,8.485 ). P(5 Y 8) P(5.5 N 8.5) (4 s.f.) c Using the inverse normal distribution, P( N a).5 a 66.4 So P( N 66.5).5 and P( N 65.5).5 So 65.5 y 66.5, i.e. the smallest value of y such that P( Y y).5 is y = The distribution satisfies the two necessary conditions for an approximation by a normal distribution to be valid: n = 8 is large (> 5) and p =.4 is close to.5. np and np( p ) (4 s.f.) SoY N(3, 4.38 ) P( X 3) P( Y 3.5) (4 s.f.) 7 a Use the binomial distribution X B(,.55) Using the binomial CD function, P( X ) P( X ) (4 s.f.) b A normal approximation is valid since n = is large (> 5) and p =.55 is close to.5. np.55 and np( p ) (4 s.f.) So Y N(, 7.36 ) P( X 95) P( Y 95.5) (3 s.f.) c It seems unlikely that the company s claim is correct: if it were true, the chance of only 95 (or fewer) seedlings producing apples from a sample of seedlings would be less than %. 8 a Use the binomial distribution X B(5,.5) Using the binomial CD function, P( X ) P( X ) (4 s.f.) b A normal approximation is valid since n = 3 is large (> 5) and p =.5 is close to.5. np and np( p ) (4 s.f.) So Y N(56,8.653 ) P( X 7) P( Y 69.5) (4 s.f.) c There is a greater than 5% chance that 7 people out of 3 would be cured, therefore there is insufficient evidence for the herbalist s claim that the new remedy is more effective than the original remedy. Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 7

8 9 X ~ N(, ) : 7, : 7, one-tailed test at the 5% level. Assume, so that Let Z X 7 5 X N(7, ) and X 7, 5 Using the inverse normal function, P( Z z).5 z.6449 X X So the critical region is X or 7.66cm (3 s.f.). Let B represent the amount of water in a bottle, so B ~ N(, ). : 5, : 5, one-tailed test at the 5% level. Assume, so that B N(5, ) and B 5, 5 Using the normal CD function, P( B 4.) (3 s.f.).67 >.5 so not significant, so accept. There is insufficient evidence to conclude that the mean content of a bottle is lower than the manufacturer s claim. Let B represent the breaking strength, so B ~ N(7.,.5 ). a Using the normal CD function, P(74.5 B 75.5) (3 s.f.) b.5 n 5 so B ~ N 7., 5 Using the normal CD function, P( B 7.4) (3 s.f.) c : 7., : 7. one-tailed test at the 5% level. Assume, so that B.5 N(7.,.5 ) and B 7., 5 (as before). Using the normal CD function, P( B 7.4) (3 s.f.).69 >.5 so not significant, so accept. There is insufficient evidence to conclude that the mean breaking strength is increased. Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 8

9 Let W represent the weight of sugar in a packet, so a W ~ N(, ). P W.95 P( W ).5 P Z.5 Using the inverse normal function, z so ( d.p.) b n 8 and x 89., so x :, :, two-tailed test with % in each tail. 6.3 Assume, so that W N(, 6.3) and W, 8 Using the normal CD function, P( W ) (3 s.f.).9 >. so not significant, so accept. There is insufficient evidence of a deviation in the mean from, so we can assume that condition i is being met. 3 Let D represent the diameter of a little-gull egg, so D ~ N(4.,.9 ). a Using the normal CD function, P(3.9 D 4.5) (4 s.f.) b.9, n 8, d 34.5, d 4.35 : 4., : 4., two-tailed test with.5% in each tail. Assume, so that.9 D ~ N(4.,.9 ) and D ~ 4., Using the normal CD function, P( D 4.35) ( s.f.).3 <.5 so significant, so reject. There is evidence that the mean diameter of eggs from this island is different from elsewhere. 4 X ~N, 8 a ~ N, X n Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 9

10 5 P 5 P Z Require P n n Z.95 P Z Using the inverse normal function, z n so n n So a sample of at least 8 is needed. 4 b X (by symmetry) Challenge a Use the binomial distribution X B(5,.48). Using the binomial CD function, P( X 8) P( X 8) (4 s.f.) b The distribution satisfies the two necessary conditions for an approximation by a normal distribution to be valid: n = 5 is large (> 5) and p =.48 is close to.5. np 5.48 and np( p ) (3 s.f.) Test to determine whether the mean is different from (the expected number of supporters based on the manager s claim): :, :, two-tailed test with.5% in each tail. 7.9 Assume, so that X ~ N(, 7.9 ) and X ~, 5 Using the inverse normal function, P( X x).5 x 4.56 and P( X x).5 x So the critical region approximated for the binomial distribution is X 5 or X 35. (Note: 5 lies in the critical region because, as part of the normal distribution, it includes the region between 4.5 and 5.5 and 4.53 is where the critical region starts. Similarly 35 lies in the critical region because, as part of the normal distribution, it includes the region between 34.5 and 35.5 and is where the critical region starts.) c Since 5, there is sufficient evidence to reject, i.e. there is sufficient evidence to say, at the 5% level, that the level of support for the manager is different from 48%. Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.