Question

In each case, determine the value of the constant c that makes the probability statement correct. (Round your answers to two decimal places.)(a) ?(c) = 0.9821(b) P(0 ? Z ? c) = 0.2939(c) P(c ? Z) = 0.1335(d) P(?c ? Z ? c) = 0.6476(e) P(c ? |Z|) = 0.0128

Answer 1

Answer:

a) c= 2.10

b) c= 0.63

c) c= -1.11

d) c= 0.93 and -c= -0.93

e) c= -0.02

Step-by-step explanation:

Hello!

The complete question is:

In each case, determine the value of the constant c that makes the probability statement correct. (Round your answers to two decimal places)

a) ∅(c)= 0.9821

b) P(0 ≤ Z ≤ c)= 0.2939

c) P(c ≤ Z) = 0.1335

d) P(-c ≤ Z ≤ c)= 0.6476

e) P( c ≤ IZI) = 0.0128

In the given probabilities is stated that you have to work under the standard normal distribution:

Z≈N (0;1)

Z= (X - μ)/(σ√n))

Normally you are given the values of a variable X and you need standardize them using the formula and look for the probabilities in the Z-table. In this case you have to do the reverse work, this means, look in the table the given probabilities and find the asociated Z number to it.

a) ∅(c)= 0.9821

Using this notation is basically as telling you that below the constant "c" there iare no values, i.e. the constat is the minimum value of the defined range of the variable,  you can also write it as:

P(Z ≤ c)= 0.9821

Since the given probability is greater than 0.50 you have to use the right Z-table and expect the constant to be positive.

c= 2.10

b) P(0 ≤ Z ≤ c)= 0.2939

This expression means that between 0 (the distribution mean) and "c" is a probability of 0.2939, an equivalent expression is:

P( Z ≤ c) - P(0 ≤ Z)= 0.2339

The probability of P(0 ≤ Z)= 0.5, so:

P( Z ≤ c) - 0.5= 0.2339

P( Z ≤ c) = 0.2339 + 0.5

P( Z ≤ c) = 0.7339

Now you look for the probability in the table, again it is greater than 0.5 so the constant is a positive number.

c= 0.63

c) P(c ≤ Z) = 0.1335

You can look for this number directly in the table, without doing any calculations. Notice that the given probability is less than 0.5, this means that the number is negative, you have to use the left Z-table.

c= -1.11

d) P(-c ≤ Z ≤ c)= 0.6476

This one is a little more tricky, it is better to resolve this graphically so I'll attach a sketch. Remember the Z table is symmbetric around cero, this means that "-c" and "c" are the same number with different sign.

Second you have to take into account is that the area unser the curve represents the max value of probability (1)

So if between -c and c is found the probability of 1 - α:0.6476, then there is α probability under the rest of the curve.

Now if you look at the graphic, there the interval between (-c;c) leaves two tails, the probability "left out" of the interval is equally disttributed in these two tails, sumbolically: α/2.

Now to the numbers:

1 - α= 0.6476

Then

α= 1 - 0.6476= 0.3524

And

α/2= 0.3524/2= 0.1762

If you look at the graphic you will se that the cumulated probabilities are:

P(Z ≤ -c) = α/2= 0.1762

And

P(Z ≤ c)) α/2 + (1-α)= 0.1762 + 0.6476= 0.8238

Now you can look in the left table for 0.1762 or in the right table for 0.8238, IcI should be the same:

c= 0.93 and -c= -0.93

e) P( c ≤ IZI) = 0.0128  

First you have to take of the module:

P(-Z ≤ c ≤ Z)= 0.0128

P(c ≤ Z) - P(c ≤ -Z)= 0.0128

P(c ≤ Z) - (1 - P(c ≤ Z))= 0.0128

2*P(c ≤ Z) - 1 = 0.0128

P(c ≤ Z)= (0.0128 + 1)/2= 0.5064

1 - P(Z ≤ c)= 0.5064

- P(Z ≤ c)= 0.5064 - 1= -0.4936

P(Z ≤ c)= 0.4936

c= -0.02

I hope it helps!

Answer 2

The probability statements follow a normal distribution.

The values of constant c are: 2.10, 0.62, -1.11, -0.93 and -0.02

(a) ∅(c)= 0.9821

The above statement means:

$\mathbf{P(z \le c) = 0.9821}$

Using z-score probability table, we have:

$\mathbf{c = 2.10}$

b) P(0 ≤ Z ≤ c)= 0.2939

The above statement means that:

$\mathbf{P( z \le c) - P(0 \le z)= 0.2339}$

Using z-score probability table, we have:

$\mathbf{ P(0 \le z)= 0.5}$

So, the equation becomes

$\mathbf{ P(z \le c)- 0.5 = 0.2339}$

Add 0.5 to both sides

$\mathbf{ P(z \le c)= 0.7339}$

Using z-score probability table, we have:

$\mathbf{c = 0.62}$

c) P(c ≤ Z) = 0.1335

$\mathbf{P( c \le z) )= 0.1335}$

Using z-score probability table, we have:

$\mathbf{c =-1.11}$

d) P(-c ≤ Z ≤ c)= 0.6476

We have:

$\mathbf{P(-c \le z \le c) = 0.6476}$

To do this, we make use of significant level

$\mathbf{P(-c \le z \le c) = P(z \le -c) = \frac{\alpha}{2}}$

Where:

$\mathbf{\alpha = 1 - 0.6476}$

$\mathbf{\alpha = 0.3524}$

So, we have:

$\mathbf{P(-c \le z \le c) = P(z \le -c) = \frac{\alpha}{2}}$

$\mathbf{P(z \le -c) = \frac{0.3524}{2}}$

$\mathbf{P(z \le -c) = 0.1762}$

Using z-score probability table, we have:

$\mathbf{c =-0.93}$

e) P( c ≤ IZI) = 0.0128

The above statement means that:

$\mathbf{P(-z \le c \le z)= 0.0128}$

Split

$\mathbf{P(c \le z) - P(c \le -z)= 0.0128}$

Using the complement rule, we have:

$\mathbf{P(c \le z) - (1 - P(c \le z))= 0.0128}$

Expand

$\mathbf{P(c \le z) - 1 + P(c \le z)= 0.0128}$

Collect like terms

$\mathbf{P(c \le z) + P(c \le z) - 1= 0.0128}$

$\mathbf{2P(c \le z) - 1= 0.0128}$

Add 1 to both sides

$\mathbf{2P(c \le z) = 1.0128}$

Divide both sides by 2

$\mathbf{P(c \le z) = 0.5064}$

Using z-score probability table, we have:

$\mathbf{c =-0.016}$

Approximate

$\mathbf{c =-0.02}$

Hence, the values of constant c are: 2.10, 0.62, -1.11, -0.93 and -0.02

Read more about normal distributions at:

https://brainly.com/question/12421652