In each case, determine the value of the constant c that makes the probability statement correct….
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Question “In each case, determine the value of the constant c that makes the probability statement correct….”
In each case, determine the value of the constant
c
that makes the probability statement correct. (Round your answers
to two decimal places.)
= 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
Normal distribution:
Normal distribution is continuous data distribution that follows a bell-shaped curve. Normally distributed random variable x has mean
Standard deviation \left( \sigma \right)
The standard normal distribution is a normal curve that has a mean of 0 and a standard deviation of 1. Thus, the parameters involved in a normal distribution are mean
.
Below is the procedure for finding the z-value
1. Find the probability value from the table of standard normal distribution.
2. Move left until you reach the first column.
3. Move up until you reach the top row.
4. The answer is equal to the sum of the column and row values.
(a)
It is evident from the information that P\left( {z \le c} \right) = 0.9821
Take a look at the table below.
Below is the procedure for finding the c-value
1. Find the probability value 0.9821 from the table of standard normal distribution.
2. Move left until you reach the first column. Take note of the value of 2.1.
3.Keep moving upwards until you reach the top row. Take note of the value as 0.00.
The area to the left z = 2.1 is created by the intersection of row and column values
Hence, P\left( {z \le c} \right) = 0.9821
(b)
It is evident from the information that P\left( {0 \le z \le c} \right) = 0.2939
Below is the procedure for finding the c-value
1. Find the probability value for 0.7939 from the table of standard normal distribution
2.Keep moving left until you reach the first column. Take note of the value at 0.8
3.Move up until you reach the top row. Take note of the 0.02 value.
The area to the left z = 0.82 is created by the intersection of row and column values
Below is a reference to the table
(c)
It is evident from the information that P\left( {c \le z} \right) = 0.1335
This is it.
\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}Below is the procedure for finding the c-value
1. Find the probability value for 0.8665.
2.Move left until you reach the first column. Take note of the value at 1.1
3.Keep moving upwards until you reach the top row. Take note of the 0.01 value.
The area to the left Z = 1.11 is created by the intersection of row and column values
Below is the graphical representation of the table:
(d)
It is evident that the total probability equals 1.
This is the probability that there is no equation:
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}Below is the procedure for finding the c-value
1. Find the probability value for 0.8238 from the table of normal standard distribution
2.Move left until you reach the first column. Take note of the $math_tag_11 value.
3.Move up until you reach the top row. Take note of the 0.03 value.
The area to the left z = 0.93 is created by the intersection of row and column values
Below is the table presentation
c are therefore -0.93 & +0.93 respectively
(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) = 0.0064Below is the procedure for finding the c-value
1. Find the probability value for 0.0064 from the table of standard normal distribution.
2. Move left until you reach the first column. Take note of the $math_tag_15 value
3.Keep moving upwards until you reach the top row. Take note of the 0.09 value.
is determined by the intersection of row and column values.
This is $math_tag_16.
The probability that z will be less or equal to c is 0.9821. This value of c is 2.1.
Part b
The value of 0.82 is the probability that z lies between 0 & c.
Part c
1.11 is the value of c for the probability that z will be greater than or equal c.
Part d
The probability of z lies between - Part E.
The probability of z lies between - and c.
Conclusion
Above is the solution for “In each case, determine the value of the constant c that makes the probability statement correct….“. We hope that you find a good answer and gain the knowledge about this topic of math.