Question

In: Math

Given a standardized a normal distribution (with a mean of 0 and a standard deviation of...

Given a standardized a normal distribution (with a mean of 0 and a standard deviation of 1, as in Table E.2), what is that probability that a. Z is less than 1.57? b. Z is greater than 1.84? c. Z is between 1.57 and 1.84? d. Z is less than 1.57 or greater than 1.84?

Solutions

Expert Solution

Concepts and reason

The concept used to solve this problem is standardization of normal distribution.

The normal distribution is used for finding the probability of the continuous variable when the data are more or less symmetric. The probability for being less than or more than some value can be calculated by calculating the area of the curve to the left of that value.

The concept of standard normal distribution can be used to calculate the probability of a normally distributed value. The random variable can be said to be follow standard normal distribution if the mean of the distribution is 0 and the standard deviation is 1.

Fundamentals

The formula to calculate the probability for a standard normal distribution is:

P(Z<z)=Φ(z)P\left( {Z < z} \right) = \Phi \left( z \right)

Here, ZZ is the standard normal variate with mean 0 and standard deviation 1.

(a)

The probability is calculated as:

P(Z<1.57)=0.9418P\left( {Z < 1.57} \right) = 0.9418

The value of the Z-score is obtained from the standard normal table.

(b)

The probability can be obtained as:

P(Z>1.84)=1P(Z1.84)=10.9671=0.0329\begin{array}{c}\\P\left( {Z > 1.84} \right) = 1 - P\left( {Z \le 1.84} \right)\\\\ = 1 - 0.9671\\\\ = 0.0329\\\end{array}

The value of the Z-score is obtained from the standard normal table.

(c)

The probability can be obtained as:

P(1.57Z1.84)=P(Z1.84)P(Z1.57)=0.96710.9418=0.0253\begin{array}{c}\\P\left( {1.57 \le Z \le 1.84} \right) = P\left( {Z \le 1.84} \right) - P\left( {Z \le 1.57} \right)\\\\ = 0.9671 - 0.9418\\\\ = 0.0253\\\end{array}

The value of the Z-score is obtained from the standard normal table.

(d)

The probability that Z is less than 1.57 and greater than 1.84 can be obtained as:

P(z<1.57andz>1.84)=1P(1.57Z1.84)=10.0253=0.9747\begin{array}{c}\\P\left( {z < 1.57{\rm{ and z > 1}}{\rm{.84}}} \right) = 1 - P\left( {1.57 \le Z \le 1.84} \right)\\\\ = 1 - 0.0253\\\\ = 0.9747\\\end{array}

The value of the Z-score is obtained from the standard normal table.

Ans: Part a

The probability for Z to be less than 1.57 is 0.94180.9418 .

Part b

The probability for Z to be greater than 1.84 is 0.03290.0329 .

Part c

The probability that Z should lie between 1.57 and 1.84 is 0.02530.0253 .

Part d

The probability that Z is less than 1.57 and greater than 1.84 is 0.97470.9747 .


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