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igomit [66]
3 years ago
7

Heights of 10-year-olds, regardless of gender, closely follow a normal distribution with a mean of 55 inches and a standard devi

ation of 6 inches.(a) What is the probability that a randomly chosen 10 year old is shorter than 48 inches?(b) What is the probability that a randomly chosen 10 year old is between 60 and 65 inches?(c) If the tallest 10% of the class is considered "very tall", what is the height cutoff for "very tall"?(d) The height requirement for Batman the Ride at Six Flags Magic Mountain is 54 inches. What percent of 10 year olds cannot go on this ride?
Mathematics
2 answers:
Svetllana [295]3 years ago
7 0

Answer:

Step-by-step explanation:

Given that Heights of 10-year-olds, regardless of gender, closely follow a normal distribution with a mean of 55 inches and a standard deviation of 6 inches.

If X is height then X is N(55,6)

Z score would be \frac{x-55}{6}

a) P(X

b) P(60

c) Tallest 10% z = 1.28

X score = 48+1.28(6) = 55.68  inches

d)P(X

i.e. 16.37% cannot go on this ride.

DENIUS [597]3 years ago
3 0

Answer:

a) 12.10% probability that a randomly chosen 10 year old is shorter than 48 inches.

b) 15.58% probability that a randomly chosen 10 year old is between 60 and 65 inches.

c) The cutoff for very tall is 62.68 inches.

d) 43.25% of 10 year olds cannot go on this ride.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 55, \sigma = 6

(a) What is the probability that a randomly chosen 10 year old is shorter than 48 inches?

This is the pvalue of Z when X = 48. So:

Z = \frac{X - \mu}{\sigma}

Z = \frac{48 - 55}{6}

Z = -1.17

Z = -1.17 has a pvalue of 0.1210.

So there is a 12.10% probability that a randomly chosen 10 year old is shorter than 48 inches.

(b) What is the probability that a randomly chosen 10 year old is between 60 and 65 inches?

This is the value of Z when X = 65 subtracted by the pvalue of Z when X = 60.

X = 65

Z = \frac{X - \mu}{\sigma}

Z = \frac{65 - 55}{6}

Z = 1.67

Z = 1.67 has a pvalue 0.9525

X = 60

Z = \frac{X - \mu}{\sigma}

Z = \frac{60 - 55}{6}

Z = 0.83

Z = 0.83 has a pvalue 0.7967

So there is a 0.9525-0.7967 = 0.1558 = 15.58% probability that a randomly chosen 10 year old is between 60 and 65 inches.

(c) If the tallest 10% of the class is considered "very tall", what is the height cutoff for "very tall"?

This is the value of X when Z has a pvalue of 0.9. So it is X when Z = 1.28

Z = \frac{X - \mu}{\sigma}

1.28 = \frac{X - 55}{6}

X - 55 = 1.28*6

X = 7.68+55

X = 62.68

The cutoff for very tall is 62.68 inches.

(d) The height requirement for Batman the Ride at Six Flags Magic Mountain is 54 inches. What percent of 10 year olds cannot go on this ride?

This is the pvalue of Z when X = 54.

Z = \frac{X - \mu}{\sigma}

Z = \frac{54 - 55}{6}

Z = -0.17

Z = -0.17 has a pvalue of 0.4325.

So 43.25% of 10 year olds cannot go on this ride.

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b)  0

c) a(t) = x''(t) = v'(t) =6t-12

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Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9

a(t) = x''(t) = v'(t) =6t-12

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

v(t) = 3t^2 -12t +9

We can factorize this function as v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is 0\leq t\leq 10 we need to analyze the following intervals:

0< t

For this case if we select two values let's say 0.25 and 0.5 we see that

v(0.25) =6.1875, v(0.5)=3.75

And we see that for a=0.5 >0.25=b we have that f(b)>f(a) so then the function is decreasing on this case.  

1

We have a minimum at t=2 since at this value w ehave the vertex of the parabola :

v_x =-\frac{b}{2a}= -\frac{-12}{2*3}= -2

And at t=-2 v(2) = -3 that represent the minimum for this function, we see that if we select two values let's say 1.5 and 1.75

v(1.75) =-2.8125< -2.25= v(1.5) so then the function sis decreasing on the interval 1<t<2

2

We see that the function would be increasing.

3

For this interval we will see that for any two points a,b with a>b we have f(a)>f(b) for example let's say a=3 and b =4

f(a=3) =0 , f(b=4) =9 , f(b)>f(a)

The particle is moving to the right then the velocity is positive so then the answer for this case is: 0

Part c

a(t) = x''(t) = v'(t) =6t-12

When the acceleration is 0 we have:

6t-12=0, t =2

And if we replace t=2 in the velocity function we got:

v(t) = 3(2)^2 -12(2) +9=-3

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