Question

Problem 7-35 A satellite range prediction error has the standard normal distribution with mean 0 NM and standard deviation 1 NM. Find the following probabilities for the prediction error: (a) less than 1.33 NM

Answer 1

Answer:

0.9082 is the probability that the prediction error is less than 1.33 NM

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 0 NM

Standard Deviation, σ = 1 NM

We are given that the distribution of prediction error is a bell shaped distribution that is a standard normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(less than 1.33 NM)

$P( x < 1.33) = P( z < \displaystyle\frac{1.33 - 0}{1}) = P(z < 1.33)$

Calculation the value from standard normal z table, we have,  

$P(x < 1.33) =0.9082= 90.82\%$

0.9082 is the probability that the prediction error is less than 1.33 NM

Answer 2

The probability for the prediction error less than 1.33 NM should be 0.9082.

Calculation of the probabilities:

since

A satellite range prediction error has the standard normal distribution with mean 0 NM and standard deviation 1 NM.

So here

P (z<1.33 - 0/1) = P(z<1.33)

So,

P(z<1.33) = 0.9082 = 90.82%

hence, The probability for the prediction error less than 1.33 NM should be 0.9082.

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