Question

Given a normal distribution with μ=55 and σ=4​, complete parts​ (a) through​ (d).

a) What is the probability that X>48​? ​(Round to four decimal places as​ needed.)

b) What is the probability that X<47​? ​(Round to four decimal places as​ needed.)

c) For this distribution, 6% of the values are less than what X-Value? ​(Round to four decimal places as​ needed.)

d) Between what two X-values (symmetrically distributed around the​ mean) are 60% of the values? ​(Round to four decimal places as​ needed.)

Answer 1

Solution :

Given that ,

mean = = 55

standard deviation = = 4

(a)

P(x > 48) = 1 - P(x < 48)

= 1 - P((x - ) / < (48 - 55) / 4)

= 1 - P(z < -1.75)

= 1 - 0.0401

= 0.9599

Probability = 0.9599

(b)

P(x < 47) = P((x - ) / < (47 - 55) / 4)

= P(z < -2)

= 0.0228

Probability = 0.0228

(c)

P(Z < z) = 6%

P(Z < -1.555) = 0.06

z = -1.555

Using z-score formula,

x = z * +

x = -1.555 * 4 + 55 = 48.78

X value = 48.78

(d)

Middle 60% has z value : -0.84 , +0.84

x = -84 * 4 + 55 = 51.64

X value = 48.78

x = 0.84 * 4 + 55 = 58.36

Two X values are :  51.64 and 58.36