Respuesta :
For the standard normal distribution Z, the mean μ =0
and σ the standard deviation = 1
a) P(0 ≤ Z ≤ 1.5) →P(Z=1.5) - P(Z=0)
P(0 ≤ Z ≤ 1.5)= 0.9332 - 0.5 = 0.4322
b) P(Z ≥ 2.5) →P(Z=2.5) = 1- P(Z=2.5)
P(Z ≥ 2.5) = 1- 0.9938 = 0.0062
c) P(Z≥ 3.5) = 1- P(Z = 3.5) = 0
OR P(Z≤ 3.5) = 0
and σ the standard deviation = 1
a) P(0 ≤ Z ≤ 1.5) →P(Z=1.5) - P(Z=0)
P(0 ≤ Z ≤ 1.5)= 0.9332 - 0.5 = 0.4322
b) P(Z ≥ 2.5) →P(Z=2.5) = 1- P(Z=2.5)
P(Z ≥ 2.5) = 1- 0.9938 = 0.0062
c) P(Z≥ 3.5) = 1- P(Z = 3.5) = 0
OR P(Z≤ 3.5) = 0
The required percentages are:
(a) 86.64%
(b) 0.62%
(c) 0.04%
With the use of standard normal table, we can find the required percentage, such as:
(a)
→ [tex]P( -1.5<z<1.5)= P( z <1.5)- p( z < -1.5)[/tex]
[tex]= 0.9332-0.0668[/tex]
[tex]= 0.8664[/tex]
[tex]= 86.64[/tex] (%)
(b)
→ [tex]P( z >2.5)=0.0062[/tex]
[tex]= 0.62[/tex] (%)
(c)
→ [tex]P( z < -3.5) + p( z > 3.5) = 0.0002+0.0002[/tex]
[tex]= 0.0004[/tex]
[tex]=0.04[/tex] (%)
Thus the above approach is right.
Learn more about standard deviation here:
https://brainly.com/question/16555520