Question

Given a normal distribution with the μ=52 and σ=3​, complete parts​ (a) through​ (d).

a. What is the probability that X>47​?

P(X>47​)=0.9525

​(Round to four decimal places as​ needed.)

b.What is the probability that X<49​?

​P(X<49​)=0.1587

​(Round to four decimal places as​ needed.)

c.For this​ distribution, 5​% of the values are less than what​ X-value?

X =

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

d. Between what two​ X-values (symmetrically distributed around the​ mean) are 80​% of the​ values?

Between what two​ X-values (symmetrically distributed around the​ mean) are 85% of the​ values?

Between what two​ X-values (symmetrically distributed around the​ mean) are 90​% of the​ values?

Between what two​ X-values (symmetrically distributed around the​ mean) are 95% of the​ values?

I need help with parts c.) and d.) please!

Answer 1

Solution :

Given that ,

mean = = 52

standard deviation = = 3

(a)

P(x > 47) = 1 - P(x < 47)

= 1 - P((x - ) /  < (47 - 52) / 3)

= 1 - P(z < -1.67)

= 1 - 0.0475   

= 0.9525

P(x > 47) = 0.9525

Probability = 0.9525

(b)

P(x < 49) = P((x - ) / < (49 - 52) / 3) = P(z < -1)

Using standard normal table,

P(x < 48) = 0.1587

Probability = 0.1587

(c)

P(Z < z) = 5% = 0.05

P(Z < -1.645) = 0.05

z = -1.645

Using z-score formula,

x = z * +

x = -1.645 * 3 + 52 = 47.065

X value = 47

(d)

(1)

P(Z < z) = 0.80

P(Z < 0.8416) = 0.80

z = +0.8416 and z = -0.8416

  x = -0.8416 * 3 + 52 = 49

x = 0.8416 * 3 + 52 = 55

X values are = 49 and 55

2)

P(Z < z) = 0.85

P(Z < 1.036) = 0.85

z = +1.036 and z = 1.036

  x = -1.036 * 3 + 52 = 49

x = 1.036 * 3 + 52 = 55

X values are = 49 and 55

3)

P(Z < z) = 0.90

P(Z < 1.28) = 0.90

z = +1.28 and z = -1.28

  x = -1.28 * 3 + 52 = 48

x = 1.28 * 3 + 52 = 56

X values are = 48 and 56

4)

P(Z < z) = 0.95

P(Z < 1.645) = 0.95

z = +1.645 and z = -1.645

  x = -1.645 * 3 + 52 = 47

x = 1.645 * 3 + 52 = 57

X values are = 47 and 57