In: Statistics and Probability

Question 1: Assume that the random variable X is normally distributed, with mean that = 47 and standard deviation that = 7. Compute the probability. Be sure to draw a normal curve with the area corresponding to the probability shaded.

P(X< AND = TO 43)

Using technology, what is P(X< AND = TO 43) equal? (round to four decimal places)

- How did you find this answer using a graphing calculator??

Question 2: The mean incubation time for a type of fertilized egg kept at 100.8°F is 22 days. Suppose that the incubation times are approximately normally distributed with a standard deviation of 2 days.

(a) What is the probability that a randomly selected fertilized egg hatches in less than 20 days?

(b) What is the probability that a randomly selected fertilized egg takes over 24 days to hatch?

(c) What is the probability that a randomly selected fertilized egg hatches between 18 and 22 days?

(d) Would it be unusual for an egg to hatch in less than 17 days? Why?

- How did you find these answers? Thanks!

Since the random variable X is normally distributed, with mean that = 47 and standard deviation that = 7 hence

to find

a) is calculated using the Z score calculator as

hence

P(X</=43)=P(Z</=-0.57)

The probability value is calculated using the Z table shown below as

P-value=0.2843

b) since the mean incubation time for a type of fertilized egg kept at 100.8 °F is 22 days. Suppose that the incubation times are approximately normally distributed with a standard deviation of 2 days.

hence,

a) P(X<20)

Using Z formula

P(Z<-1)=0.1587 using Z table

b) P(X.>24)

P(Z>1)=0.1587 again computed using the Z table shown below

c) P(18<X<22)

Z at X=18

and Z at X=22

So, P(10<X<22)

=P(-2<Z<0)

=0.50-0.0228

=0.4772

d) Again P(X<17)

According to Central limit theorem if Z value is less than -2 and greater than 2 then by rare event rule the probability of happening that event becomes very low hence it is unusual.

Z table

Assume that the random variable x is normally
distributed with mean μ = 70 and standard deviation
σ = 10. Find P(x < 82)
a. 0.9942
b. 0.4500
c. 0.1151
d. 0.8849

Assume the random variable X is normally distributed with mean μ
= 50 and standard deviation σ = 7. Compute the probability.
P(35<X<63)

Assume that the random variable X normally distributed, with mean
90 and standard deviation 15. Compute the probability
P(X>102).

Let x be a continuous random variable
that is normally distributed with a mean of 65 and a standard
deviation of 15. Find the probability that
x assumes a value:
less than 48
greater than 87
between 56 and 70

5. X is a normally distributed random variable with a mean of 8
and a standard deviation of 3. The probability that X is between 6
and 10 is
a. 0.7486
b. 0.4972
c. 0.6826
d. 0.8413
The weight of football players is normally distributed with a
mean of 200 pounds and a standard deviation of 20 pounds.
6. The probability of a player weighing more than 240 pounds
is
a. 0.0197
b. 0.9803
c. 0.4803
d. 0.0228
7. Refer...

Suppose that y = x2, where x is a normally distributed random variable with a mean and variance of µx = 0 and σ2x = 4. Find the mean and variance of y by simulation. Does µy = µ2x? Does σy = σ2x? Do this for 100, 1000, and 5000 trials.

A random variable X is Normally distributed with mean = 75 and =
8. Let Y be a second Normally distributed random variable with mean
= 70 and = 12. It is also known that X and Y are independent of one
another. Let W be a random variable that is the difference between
X and Y (i.e., W = X – Y). What can be said about the distribution
of W?

1)If a variable X is normally
distributed with a mean of 60 and a standard deviation of 15, what
is
P(X > 60)
P(X > 75)
P(X > 80)
P(X < 50)
P(45 < X < 75)
P(X < 45)

Suppose a random variable X is normally distributed
with mean 65.9 and standard deviation 9.5. Answer the following
questions:
P(42.15 < X < 77.30) = [round to 4
decimal places]
Tries 0/5
P(X ≤ 73.50) = [round to 4 decimal
places]
Tries 0/5
P(X = 77.30) = [round to 4 decimal
places]
Tries 0/5
Suppose a is such that: P(X ≤
a) = 0.49. Then a = [round to 2 decimal
places]
Tries 0/5
What is the IQR (inter-quartle range)...

Suppose a random variable X is normally distributed with mean
65.3 and standard deviation 9.4. Answer the following
questions:
1. P(47.44 < X < 76.58) = [round to 4 decimal places]
2. P(X ≤ 46.50) = [round to 4 decimal places]
3. P(X = 76.58) = [round to 4 decimal places]
4. Suppose a is such that: P(X ≤ a) = 0.39. Then a = [round to 2
decimal places]
5. What is the IQR (inter-quartle range) of X? [round...

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