SOLUTION: 1.Suppose the time to transmit an email message is normally distributed with a mean of 0.72 seconds and a standard deviation of 0.10 seconds. a.What is the probability that a

Question 969825: 1.Suppose the time to transmit an email message is normally distributed with a mean
of 0.72 seconds and a standard deviation of 0.10 seconds.
a.What is the probability that an email message will require
i.more than one second to transmit
ii.between 0.70 and 0.80 second to transmit?
b.90% of the messages will take at least how many seconds to be transmitted?
Suppose that a sample of 16 email messages is selected.
c.What is the probability that the average time for an email message to be transmitted is
i.more than 0.7 second?
ii.Between 0.7 and 0.8 second?
d.90% of the average transmission times will be symmetrically between what two values?

2.The average percentage of brown M&M candies in a package of plain M&Ms is 30%. Suppose you randomly select a package of plain M&Ms that contains 55 candies and determine the proportion of brown candies in the package.
a.What is the probability that the sample proportion of brown candies is less than 20%?
b.What is the probability that the sample proportion of brown candies is not within 3.5% of the population proportion?
c.What is the probability that the sample proportion exceeds 35%?
d.Within what range would you expect the sample proportion to lie symmetrically about 80% of the time?
*my friends and I had different problems so that I combine the questions that we can't solves this problems. Please help us thank you!
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
mean= 0.72 s sd=0.10 s
z= (1-0.72)/0.10= 0.28/0.10 and it is a one sided z-score of +2.8
P (z>2.8)=0.0026
ii. That would be a z-score of (0.70-0.72)/.10 or z-score of -0.2 P=0.0793
That would be a z-score of (0.8-0.72)/10 or z-score of + 0.8 P=0.2881 ; Total P=0.3674
b. 90% messages: would be a z-score of >+1.28 or .128 sec or 0.848 (0.85) sec
c. Sampling distribution
Probability z < 0.70 sec: if we get that, we subtract it from 1
z score is- 0.02/[(0.1)/4] = 0.08/1 z=-0.8 That probability is 0.2881, so probability > 0.70 sec in sampling distribution is 1-0.2881= 0.7119.
Between 0.7 sec and 0.8 sec
(0.7-0.72)/(0.1/4); that is - 0.08/0.1 or z>-0.8 This is 0.2881 on the left side of the curve
(0.8-0.72)/(0.1/4); that is +0.08*4/0.1 or z<3.2 This is 0.5000 on the right side
The probability is 0.7881.
90% of the average transmission time will be between z values of +/- 1.645
xb +/- 1.645 (0.1)/4 , where numerical value is 0.041
Rounding, 90% are between 0.68 and 0.76 for the sample.
++++++++++++++++++++++++++++++++++++++++++++++++
2.a z= (pb-p)/sqrt [(p*(1-p)/55]
z=(0.2-0.3)/sqrt [(.3*.7)/55]
-(0.1)/0.0618 =-1.618 This is about 0.055.
Not within 3.5% means that sample proportion is >0.335 or <0.265
Same denominator, numerator is 0.035
z value > or < 0.566 This is 0.2857 x2=0.5714
Same denominator, numerator is now 0.05 and one-way test shows probability of 0.21
80% of the time would be a z-value of +/-1.28 ; I multiply that by the denominator 0.0618 and get 0.079 or about 0.08 The interval would be [0.22, 0.38]
Note: This problem is complicated by the fact that n is small; >35% of the population is 19.25, so greater than 35% starts with 20, which is really 36.3%. There are issues with this throughout the problem. If you do it on a calculator, with z-scores, you have no problem, but if you input data that would be required to answer the question, you will get different values.
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This was a long pair of problems. I wouldn't be surprised if I missed something in this, but trying things a couple of ways, by hand and by calculator, I got things to make sense.

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