wutao is going on vacation in Hawaii! Once there, he has decided to jump over a

Question

wutao is going on vacation in Hawaii! Once there, he has decided to jump over a shark on water skis. He consulted a team of a Purdue Engineers. The team leader estimates the distance of Wutao's water skiing jumps has a Normal distribution with a mean of 24.5 feet and variance of 9.61. a) What is the 10th percentile of Wutao's watersking jumps? b) The shark is contained in a giant underwater cage that is 18 feet long. We will assume that Wutao starts his 3. jump at the beginning of the cage, what is the probability Wutao safely jumps over the shark cage? Wutao decides to practice jumping on water skis before an actual shark is involved. Given his practice jump is at least 19.8 feet, what is the probability he jumps at most 27 feet? The day of the event, there is a strong wind coming in. of his water ski jump. The extra amount that the wind will add to Wutao's jump is independently and normally distributed with a mean of 3 feet and a variance of 1 foot. What is the distribution and parameters of the length of Wutao's jumps with the addition of the wind? Under the wind conditions, what is the probability Wutao successfully jumps the shark? c) d) The wind is at Wutao's back and will add to the length e)

Explanation / Answer

Solution:

a.

We have to find the value X for the 10thpercentile.

Formula for value X is given as below:

X = Mean + Z*SD

We are given that the variable follows a normal distribution.

Mean = 24.5

Variance = 9.61

So, SD = sqrt(9.61) = 3.1

For 10th percentile, critical Z value is -1.28155

X = 24.5 + (-1.28155)*3.1

X = 20.5272

Required answer: 10th percentile = 20.5272

b.

We have to find P(X>18)

P(X>18) = 1 – P(X<18)

Z = (X – mean) / SD

We are given that the variable follows a normal distribution.

Mean = 24.5

Variance = 9.61

So, we have

Z = (18 – 24.5) / 3.1

Z = -2.09677

P(Z< -2.09677) = P(X<18) = 0.018007

P(X>18) = 1 – P(X<18)

P(X>18) = 1 – 0.018007

P(X>18) = 0.981993

Required probability = 0.981993

c.

Here, we have to find P(19.8<X<27)

P(19.8<X<27) = P(X<27) – P(X<19.8)

Z for X<27

Z = (27 - 24.5) / 3.1

Z = 0.806452

P(Z<0.806452) = P(X<27) = 0.790009

Z for X<19.8)

Z = (19.8 - 24.5) / 3.1

Z = -1.51613

P(Z<-1.51613) = P(X<19.8) = 0.064743

P(19.8<X<27) = P(X<27) – P(X<19.8)

P(19.8<X<27) = 0.790009 - 0.064743

P(19.8<X<27) = 0.725266

Required probability = 0.725266

d.

We are given X follows N(µ = 24.5, 2 = 9.61)

Wind Y follows N(µ = 3, 2 = 1)

So, X+Y follows N(µ1 + µ2, 1^2 + 2^2)

X + Y follows N(24.5 + 3, 9.61+1) = N(27.5, 10.61)

e.

We have to find P(X>18) with mean = 27.5 and variance = 10.61

SD = sqrt(10.61) = 3.257299

P(X>18) = 1 – P(X<18)

Z = (18 – 27.5) / 10.61

Z = -0.89538

P(Z< -0.89538) = P(X<18) = 0.185292

P(X>18) = 1 – P(X<18)

P(X>18) = 1 – 0.185292

P(X>18) = 0.814708

Required probability = 0.814708

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