Question

Suppose a random variable X is normally distributed with mean 65 and standard deviation 7. Answer the following questions:

1. P(48.90 < X < 70.60) = ____ [round to 4 decimal places]

2. P(X ≤ 65.70) = _____ [round to 4 decimal places]

3. P(X = 70.60) = ____ [round to 4 decimal places]

4.Suppose a is such that: P(X ≤ a) = 0.46. Then a = ____ [round to 2 decimal places]

5. What is the IQR (inter-quartle range) of X? _____ [round to 2 decimal places]

Answer 1

Solution :

Given that,

mean = = 65

standard deviation = = 7

1 ) P (48.90 < x < 70.60 )

P ( 48.90 - 65/ 7) < ( x -  / ) < ( 70.60 - 65 / 7)

P ( -16.1 / 7 < z < 5.6 / 7 )

P (-2.3 < z < -0.8 )

P ( z < 0.8 ) - P ( z < -2.3 )

Using z table

= 0.7881 - 0.0107

= 0.7774

Probability = 0.7774

2 ) P( x 65.70 )

P ( x - / ) ( 65.70 - 65 / 7)

P ( z 0.70 / 7 )

P ( z 0.1)

= 0.5398

Probability = 0.5398

3 ) P(X = 70.60) = 0

4 ) P( X a) = 0.46

a = -0.10

Using z-score formula,

x = z * +

x = -0.10 * 7 + 65

x = 64.3

5 ) P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.6745 ) = 0.25

z = -0.67

Using z-score formula,

x = z * +

x = -0.67 * 7 + 65

x = 60.31

First quartile =Q1 = 60.31

The z dist'n Third quartile is,

P(Z < z) = 75%

= P(Z < z) = 0.75  

= P(Z < 0.6745 ) = 0.75

z = 0.6745

Using z-score formula,

x = z * +

x = 0.67 * 7 + 65

x = 69.69

Third quartile =Q3 = 69.69

IQR = Q3 - Q1

= 69.69 - 60.31

= 9.38

IQR = 9.38