Question

Given a normal distribution with μ = 48 and σ = 5​,

a. What is the probability that X > 42​? ​(Round to four decimal places as​ needed.)

b. What is the probability that X < 43​? ​(Round to four decimal places as​ needed.)

c. For this​ distribution, 9​% of the values are less than what​ X-value? ​(Round to the nearest integer as​ needed.)

d. Between what two​ X-values (symmetrically distributed around the​ mean) are 60​% of the​ values?

​(Round to the nearest integer as​ needed.)

Answer 1

Solution :

Given that ,

mean = = 48

standard deviation = = 5

a) P(x > 42) = 1 - p( x< 42)

=1- p P[(x - ) / < (42 - 48) / 5]

=1- P(z < -1.20)

Using z table,

= 1 - 0.1151

= 0.8849

b) P(x < 43) = P[(x - ) / < (43 - 48) / 5]

= P(z < -1.00)

Using z table,

= 0.8413

c) Using standard normal table,

P(Z < z) = 9%

= P(Z < z) = 0.09  

= P(Z < -1.34) = 0.09

z = -1.34

Using z-score formula,

x = z * +

x = -1.34 * 5 + 48

x = 41

d) Using standard normal table,

P( -z < Z < z) = 60%

= P(Z < z) - P(Z <-z ) = 0.60

= 2P(Z < z) - 1 = 0.60

= 2P(Z < z) = 1 + 0.60

= P(Z < z) = 1.60 / 2

= P(Z < z) = 0.80

= P(Z < 0.84) = 0.80

= z  ± 0.84

Using z-score formula,

x = z * +

x = -0.84 * 5 + 48

x = 44

Using z-score formula,

x = z * +

x = 0.84 * 5 + 48

x = 52

The 60% are x = 44 and x = 52