Question

Suppose a random variable X is normally distributed with mean 69.8 and standard deviation 8. Answer the following questions:

P(55.40 <  X < 74.60) = [round to 4 decimal places]

Tries 0/5

P(X ≤ 62.60) = [round to 4 decimal places]

Tries 0/5

P(X = 74.60) = [round to 4 decimal places]

Tries 0/5

Suppose a is such that: P(X ≤ a) = 0.51. Then a = [round to 2 decimal places]

Tries 0/5

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

Answer 1

Solution:

We are given

µ = 69.8

σ = 8

Question 1

We have to find P(55.40 <  X < 74.60) =

P(55.40 <  X < 74.60) = P(X<74.60) – P(X<55.40)

First find P(X<74.60)

Z = (X - µ)/σ

Z = (74.60 - 69.8)/8

Z = 0.6

P(Z<0.6) = P(X<74.60) = 0.725747

(by using z-table)

Now find P(X<55.40)

Z = (X - µ)/σ

Z = (55.40 - 69.8)/8

Z = -1.8

P(Z< -1.8) = P(X<55.40) = 0.03593

(by using z-table)

P(55.40 <  X < 74.60) = P(X<74.60) – P(X<55.40)

P(55.40 <  X < 74.60) = 0.725747 - 0.03593

P(55.40 <  X < 74.60) = 0.689817

Required probability = 0.6898

Question 2

Here, we have to find P(X≤62.60)

Z = (X - µ)/σ

Z = (62.60 - 69.8)/8

Z = -0.9

P(Z< -0.9) = P(X≤62.60) = 0.18406

(by using z-table)

Required probability = 0.1841

Question 3

P(X=74.60) = P(X<74.60+0.5) - P(X<74.60 - 0.5) (by using continuity correction)

P(X=74.60) = P(X<75.10) – P(X<74.10)

Z = (X - µ)/σ

Z = (75.10 - 69.8)/8

Z =0.6625

P(Z<0.6625) = P(X<75.10) = 0.746175

(by using z-table)

Z = (X - µ)/σ

Z = (74.10 - 69.8)/8

Z =0.5375

P(Z<0.5375) = P(X<74.10) = 0.704539

(by using z-table)

P(X=74.60) = 0.746175 - 0.704539

P(X=74.60) = 0.041636

Required probability = 0.0416

Question 4

P(X≤a) = 0.51

So, Z = 0.025069

(by using z-table)

a = µ + Z*σ

a = 69.8 + 0.025069*8

a = 70.00055

a = 70.00

Question 5

IQR = Q3 – Q1

Z for Q3 = 0.67449

Z for Q1 = -0.67449

Q3 = µ + Z*σ = 69.8 + 0.67449*8 = 75.19592

Q1 = µ + Z*σ = 69.8 - 0.67449*8 = 64.40408

IQR = Q3 – Q1

IQR = 75.19592 - 64.40408

IQR = 10.79184

IQR = 10.79