156k views
3 votes
If a Variable has a normal distribution with mean 30 and standard deviation 5 find the probability that the variable will be between 25 and 37.

A. 0.77
B. 80
0.72
0.75

User RomanKonz
by
4.2k points

1 Answer

7 votes

Answer:

P(25 < x < 37) = 0.77

Explanation:

Given - If a Variable has a normal distribution with mean 30 and standard deviation 5

To find - find the probability that the variable will be between 25 and 37.

Proof -

Given that,

Mean, μ = 30

S.D, σ = 5

Now,


z = (x-\mu)/(\sigma) ~ N(0,1)

Now,

P(25 < x < 37)

=
P((25 - 30)/(5) < z < (37 - 30)/(5) )

= P(1 < z < 1.4)

= P(z < 1.4) - P(z < -1)

= 0.9192 - 0.1587

= 0.7605

≈ 0.77

∴ we get

P(25 < x < 37) = 0.77

User Reveille
by
4.5k points