Q1 suppose the waiting time at a bank tellers window is...

(Q:1) Suppose the waiting time at a bank teller's window is normally distributed with mean μ = 5 minutes and standard deviation σ = 1.2 minutes. The probability that a customer has to wait more than 7 minutes on a visit to the bank is ? Which one is correct answer : 0.0015 /0.9522/ 0.0478/ 0.0250 Q:2) Based on the records for millions of homeowners, an insurance company determines that the distribution of the loss from fire in a year has a mean of μ=$ 250 per house and that the standard deviation of loss is σ=$1000. Subsequently, the company sells 10,000 fire insurance policies. What is the approximate probability that for this sample of 10,000 policies tha average loss per policy in a year will be greater than $275? Which one is correct answer: 0.9938/ 0.0062/ 0.4900/ 0.5100

Transcript

00:01 Okay, you have two questions here, part a and b.

00:03 Part a is waiting time in a bank, and we are told that the mean for that is five minutes with the standard deviation of 1 .2, and we're told that it's normally, that waiting time is normally distributed.

00:19 Okay, and you're trying to find out what's the probability that a randomly selected person would have a waiting time greater than seven minutes.

00:27 I've drawn out the normal distribution, and what we're looking for here is the area under the normal curve to the right, which would represent that probability.

00:39 We could use technology for this.

00:41 We could use a z table for this.

00:44 Let me just show that real quick.

00:48 You'd have to find a z score for 7.

00:50 So it would be z equals 7 minus 5 divided by 1 .2, and this works out to be 1 .6 repeating.

00:59 Now, on a z table, i can only be accurate to two decimal places.

01:04 So there's 1 .6.

01:06 There's the second six.

01:08 Our answer is going to be between these two, and i don't think we can be real accurate.

01:14 We could probably be close enough on a z table for that value.

01:19 But looking at your answer choices, they are out to four decimal places.

01:24 And given that we have a repeating z score, i'm assuming that you would use technology for this answer.

01:29 But our answer is going to be somewhere.

01:33 Let's just say this is 0 .952.

01:38 Our answer is going to be close to 0 .048.

01:46 Now let's see if we can use technology to find that a little quicker.

01:51 The program that you should use would be normal cdf.

01:57 That's typically the name of this program.

01:59 I'm using a t .i .84.

02:02 This is found under a second distribution key.

02:04 The first thing we do is type in the lower boundary which is seven then our upper boundary which is headed off in this direction to basically infinity so you just put in a big number i like to put it a million then we put in our mean five and our standard deviation 1 .2 and using technology i'll get the answer 0478 0 .0478 so using a a z table, i had 0 .048, but technology got me a little bit more accurate, 0 .0478, which i believe was one of your answer choices.

03:05 Yes, it was the third answer choice you were given.

03:10 Okay, moving on to the second problem...

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