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For a standard normal distribution, find the approximate value of P(-0.78 = z = 1.16). Use the portion of the standard normal table below to help answer the question.

(z)
0.00
0.16
0.22
0.78
1.00
1.16
1.78
2.00

(Probabilities)
0.5000
0.5636
0.5871
0.7823
0.8413
0.8770
0.9625
0.9772

Respuesta :

asadullahgiki

Correction:

I think the question should be like that find the approximate value of

P(-0.78 < z < 1.16)?


Now,

the symbol Ф represent the cumulative density.

first find the

Ф(1.16) from the above given table which is equal to 0.8770.

Now,

find the Ф(-0.78) .

in our table we are given the value of Ф(0.78)=0.7823

so as the curve is symmetrical Ф(-0.78)=1-0.7823=0.2177


P(-0.78 < z < 1.16) = Ф(1.16)-Ф(-0.78)

= 0.8770-0.2177

= 0.6593

MrRoyal

The approximated value of P(0.78< z < 1.16) is 0.0947

The standard normal expression is given as:

[tex]P(0.78 < z < 1.16)[/tex]

To determine the value of the expression, we make use of:

[tex]P(z_1< z < z_2) = P(z < z_2) - P(z

So, we have:

[tex]P(0.78< z < 1.16) = P(z < 1.16) - P(z < 0.78)[/tex]

From the portion of the standard normal table provided, we have the following probability values

  • [tex]P(z < 1.16) = 0.8770[/tex]
  • [tex]P(z < 0.78) = 0.7823[/tex]

So, we have:

[tex]P(0.78< z < 1.16) = 0.8770 - 0.7823[/tex]

Calculate the differences

[tex]P(0.78< z < 1.16) = 0.0947[/tex]

Hence, the approximated value of P(0.78< z < 1.16) is 0.0947

Read more about standard normal distribution at:

https://brainly.com/question/18761122

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