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

Concepts and reason

Normal distribution:

Normal distribution is a continuous distribution of data that has the bell-shaped curve. The normally distributed random variable x has meanand standard deviation $\left( \sigma \right)$ .

Also, the standard normal distribution represents a normal curve with mean 0 and standard deviation 1. Thus, the parameters involved in a normal distribution are mean and standard deviation.

Fundamentals

Procedure for finding the z-value is listed below:

1.From the table of standard normal distribution, locate the probability value.

2.Move left until the first column is reached.

3.Move upward until the top row is reached.

4.The sum of row value and the column value is the corresponding answer.

(a)

From the information, it is clear that, $P\left( {z \le c} \right) = 0.9821$

Consider the normal table as shown below:

Procedure for finding the c-value is listed below:

1.From the table of standard normal distribution, locate the probability value of 0.9821.

2.Move left until the first column is reached. Note the value as 2.1.

3.Move upward until the top row is reached. Note the value as 0.00.

The intersection of the row and column values gives the area to the left of $z = 2.1$ as 0.9821.

Hence, $P\left( {z \le c} \right) = 0.9821$

(b)

From the information, it is clear that, $P\left( {0 \le z \le c} \right) = 0.2939$

Procedure for finding the c-value is listed below:

1.From the table of standard normal distribution, locate the probability value of 0.7939

2.Move left until the first column is reached. Note the value as 0.8

3.Move upward until the top row is reached. Note the value as 0.02

The intersection of the row and column values gives the area to the left of $z = 0.82$ as 0.7939.

The following is reference of the table

(c)

From the information, it is clear that, $P\left( {c \le z} \right) = 0.1335$

That is,

$\begin{array}{c}\\P\left( {z \ge c} \right) = 0.1335\\\\1 - P\left( {z \le c} \right) = 0.1335\\\\P\left( {z \le c} \right) = 1 - 0.1335\\\\ = 0.8665\\\end{array}$

Procedure for finding the c-value is listed below:

1.From the table of standard normal distribution, locate the probability value of 0.8665.

2.Move left until the first column is reached. Note the value as 1.1.

3.Move upward until the top row is reached. Note the value as 0.01.

The intersection of the row and column values gives the area to the left of $Z = 1.11$ as 0.8665.

The graphical presentation of the table is shown below:

(d)

It is clear that the total probability equals 1.

The probability that lies outside the equation is as follows:

Consider,

$\begin{array}{c}\\P\left( { - c < z < c} \right) = 0.6476\\\\P\left( { - c \le z \le 0} \right) + P\left( {0 \le z \le c} \right) = 0.6476\\\\2P\left( {0 \le z \le c} \right) = 0.6476\left( {{\rm{Since Normal distribution is symmetric}}} \right)\\\\P\left( {0 \le z \le c} \right) = 0.3238\\\end{array}$

$\begin{array}{c}\\P\left( {z \le c} \right) - P\left( {Z \le 0} \right) = 0.3238\\\\P\left( {z \le c} \right) - 0.5 = 0.3238\\\\P\left( {z \le c} \right) = 0.8238\\\end{array}$

Procedure for finding the c-value is listed below:

1.From the table of standard normal distribution, locate the probability value of 0.8238

2.Move left until the first column is reached. Note the value as $0.9$ .

3.Move upward until the top row is reached. Note the value as 0.03.

The intersection of the row and column values gives the area to the left of $z = 0.93$ .

The presentation of table is shown below:

Hence, the values of c are –0.93 and +0.93

(e)

Consider,

$\begin{array}{c}\\P\left( {z \le c} \right) - P\left( {z \le - c} \right) = 0.9872\\\\P\left( {z \le c} \right) - \left[ {1 - P\left( {z \le c} \right)} \right] = 0.9872\\\\2P\left( {z \le c} \right) = 1 - 0.9872\\\\P\left( {z \le c} \right) = \frac{{0.0128}}{2}\\\end{array}$

$P\left( {z \le c} \right) = 0.0064$

Procedure for finding the c-value is listed below:

1.From the table of standard normal distribution, locate the probability value of 0.0064.

2.Move left until the first column is reached. Note the value as $- 2.4$

3.Move upward until the top row is reached. Note the value as 0.09

The intersection of the row and column values gives the area to the left of$z=-2.49 $.

That is, $P\left( {z \le - 2.49} \right) = 0.0064$ . Hence, $P\left( { - 2.49 < z < 2.49} \right) = 0.0128$

Ans: Part a

The value of c for which probability of z less than or equal to c to be 0.9821 is 2.1.

Part b

The value of c, for the probability of z that lies between 0 and c to be 0.2939 is 0.82.

Part c

The value of c, for the probability of z greater than or equal to c to be 0.1335 is 1.11.

Part d

The values of c for the probability of z that lies between $-$ c and c to be 0.6476 is $\pm 0.93$ .

Part e

The value of c is for the probability of z that lies between $-$ c and c to be 0.0128 is 2.49.