Question

if x is a normal variance with mean 30 and standard deviation 5 find the probability that 26<=X<=40 X>=45 f(0.8)=0.2881,f(2.0)=0.4772,f(3.0)=0.4987

Answer 1

Answer:

i)for 26<=X<=40 , ans is 0.7653

ii)for  X>=45, ans is 0.0013

Step-by-step explanation:

Here mean μ= 30 and standard deviation σ = 5

(i) When X = 26, Z = ( X − µ ) / σ = (26 – 30)/5 = –4/5= -0.8

And when X = 40 , Z = [40 – 30] /5 =10/5= 2

now, forP(26<X<40) ) = P(–0.8 ≤ Z < 2)

= P(–0.8 ≤ Z ≤ 0) + p(0 ≤ Z ≤ 2)

= P(0 ≤ Z ≤ 0.8) + P(0 ≤ Z ≤ 2)

= 0.2881 + 0.4772 (given and from table)

= 0.7653

(ii) The probability that X≥45

When X = 45

Z = [X − µ] / σ = [45 – 30] / 5 = 3

=P( X ≥ 45) = P(Z ≥ 3)  

= 0.5 – 0.4987

= 0.0013