One thousand independent rolls of a fair die will be made. compute an approximation to the probability that the number 6 will appear between 150 and 200 times inclusively. if the number 6 appears exactly 200 times, find the probability that the number 5 will appear less than 150 times.

The probability that the number 6 will appear between 150 and 200 times inclusively is 0.9258

The probability that the number 5 will appear less than 150 times is 0.1762.

Solution:

Given, One thousand independent rolls of a fair die will be made.

First, let’s compute an approximation to the probability that number 6 will appear between 150 and 200 times.  

The die is rolled 1,000 times. These are independent trials which are Bernoulli trials since a die shows either a 6 or not.  

Thus, n = 1000. The probability of success is constant, p = 1/6. We can use the binomial model.  

Let X be the number of times the die shows a 6.  

We can use the normal approximation to the binomial model  

Because np = 1000 • 1/6 > 10 and n(1 − p) = 1000 • 5 6 > 10. x ∼ Bin(166.7, 11.79).

But, using the normal approximation, let Y be normally distributed using the parameters from the binomial model.

That is, in the normal model

So, let y ∼ N(166.7, 11.79).

There is one small complication. We are using a continuous distribution to model a discrete distribution. We must correct for this discrepancy using the continuity correction which is bolded in the equation below.  

P(150 ≤ Y ≤ 200) = P(Y ≤ 200) − P(Y ≤ 150)  

Now, we consider the probability that we obtain less than 150 6s if exactly 200 5s have been shown.

This is a conditional probability problem. Let Y be the number of 6s that appear and let X be the number of 5s that appear.  

Then we need to compute P(Y < 150|X = 200).  

That is, we consider that 200 of the rolls show a 5, therefore, we have 800 trials remaining that can display anything except a 5, so n = 800 now.  

Since we have already considered those 200 trials where a 5 appears, the probability of success (a 6 appears) is p = 1/5.  

Again, we use the normal approximation due to the large value for n and since the success/failure conditions hold (np > 10 and n(1 − p) > 10).  

Let Y ∼ N(160, 11.31).

Hence, the probabilities are 0.9258 and 0.1762.

2 & 7

step-by-step explanation:

he climbs 300 ft/hr at a constant rate. the peak is at 1,500 ft.

when going up, he reaches 600 ft in 2 hours. remember, that he is going in a constant rate. divide 600 with 300:

600/300 = 2

assuming that he takes the same route back (meaning that it's a 1500 ft route back, you can solve.

note also that you are solving when he gets back to 600 ft after he reaches the peak.

first, find the amount of hours it takes to get up the peak. divide the total amount (1500), with the ft/her (300) to get the amount of hours (x):

1500/300 = x

x = 5

it takes jamie 5 hours to get up. now, he is returning down, and reaches 600 (his current destination). remember that he climbs at a constant 300 ft/hr. divide 600 with 300:

600/300 = x

x = 2

now add the two numbers together.

x = 5 + 2 = 7

jamie takes 7 hours to go up the mountain and reach his current position when descending.

(total amount/constant rate) = (time).

remember that it is a constant rate, so in total, when he gets back down the mountain, it will take the same amount of time as when he goes up. the total amount for the round trip is: 5 + 5 = 10.

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