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6. A calculus exam had a mean of 75 and a standard deviation of 6. If Sally had made a 78 on this exam, what
6. A calculus exam had a mean of 75 and a standard deviation of 6. If Sally had made a 78 on this exam, what is her z-score? 16. Find the following probabilities: b. P1.68 Z 2.09 c. PZ 0.56 d. P0.52 Z 0.52 e. P2.09 Z 1.32 f. PZ 1.08 a. P Z 1.82 18. Find the value of c that satisfies each of the following probabilities: 0.0495 b. PZ c0.8438 c. PZ c0.0778 d. Pc Z c0.9774 a. P Z c 4.4 2. The mean TOEFL score of international students at a certain university is normally distributed with a mean of 490 and a standard deviation of 80. Suppose groups of 30 students are studied. Find the mean and the standard deviation for the distribution of sample means. 5. A waiter estimates that his average tip per table is $20 with a standard deviation of $4. If we take samples of 9 tables at a time, calculate the following probabilities when the tip per table is normally distributed. a. What is the probability that the average tip for one table is less than $21? b. What is the probability that the average tip for one table is more than $21? c. What is the probability that the average tip for one table is between $19 and $21? 12. In a large population, 76% of the households own microwaves. A simple random sample of 100 households is to be contacted and the sample proportion computed. What is the mean and standard deviation of the sampling distribution of the sample proportions? 6. A calculus exam had a mean of 75 and a standard deviation of 6. If Sally had made a 78 on this exam, what is her z-score =4, =75 X=78 Z= X Z= 7875 =0.5 6 16. (Note all the Probabilities are checked from Z table) a. P(Z<-1.82)=0.0344 ( b. P -1.68 0.56)=1-P(Z0.56) =1-0 .7123 = 0.2877 d. P(-0.52-1.08)=1-P(Z<-1.08) =1- 0.1401 =0.8599 18. (Note all the c values are checked from Z table) a. P(Zc)= 0.0778 P(Z21)=P( Z> 2120 4 /9 ) =P(Z>0.75) =1- P(Z<0.75) =1-0.7734 =0.2266 c. What is the probability that the average tip for one table is between $19 and $21? P(190.56)=1-P(Z0.56) =1-0 .7123 = 0.2877 d. P(-0.52-1.08)=1-P(Z<-1.08) =1- 0.1401 =0.8599 18. (Note all the c values are checked from Z table) a. P(Zc)= 0.0778 P(Z21)=P( Z> 2120 4 /9 ) =P(Z>0.75) =1- P(Z<0.75) =1-0.7734 =0.2266 c. What is the probability that the average tip for one table is between $19 and $21? P(190.56)=1-P(Z0.56) =1-0 .7123 = 0.2877 d. P(-0.52-1.08)=1-P(Z<-1.08) =1- 0.1401 =0.8599 18. (Note all the c values are checked from Z table) a. P(Zc)= 0.0778 P(Z21)=P( Z> 2120 4 /9 ) =P(Z>0.75) =1- P(Z<0.75) =1-0.7734 =0.2266 c. What is the probability that the average tip for one table is between $19 and $21? P(19
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