Question

A random sample of size n₁ = 16 is selected from a normal population with a mean of 74 and a standard deviation of 9. A second random sample of size n₂ = 7 is taken from another normal population with mean 68 and standard deviation 11. Let X1and X2 be the two sample means. Find:

(a) The probability that X1-X2 exceeds 4.

(b) The probability that 4.8 ≤X1-X2≤ 5.6.

Round your answers to two decimal places (e.g. 98.76).

Answer 1

SOLUTION:

From given data,

A random sample of size n₁ = 16 is selected from a normal population with a mean of 74 and a standard deviation of 9. A second random sample of size n₂ = 7 is taken from another normal population with mean 68 and standard deviation 11.

Let X₁ and X₂ be the two sample means.

Where,

sample size = n₁ = 16

Population with a mean = = 74

Standard deviation = = 9

sample size = n₂ = 7

Population with a mean = = 68

Standard deviation = = 11

The linear combination of independent normal random variables follows normal distribution.

The sampling distribution of X₁-X₂ is normal with mean

= -

= -

= 74 - 68

= 6

The sampling distribution of variance X₁-X₂ is,

= +

= /   + /

= 9² / 16 + 11²​​​​​​​ / 7​​​​​​​

= 22.3482

The sampling distribution of standard deviation is,

= sqrt ()

= sqrt (22.3482)

= 4.72738

(a) The probability that X₁-X₂ exceeds 4.

P(X₁-X₂ > 4) = 1 - P(X₁-X₂ ​​​​​​​ < 4)

P(X₁-X₂ > 4) = 1 - P(((X₁-X₂ )-​​​​​​​ ) / ()  < (4-​​​​​​​ 6) /4.72738 )

P(X₁-X₂ > 4) = 1 - P(Z < -0.42)

P(X₁-X₂ > 4) = 1 - 0.33724

P(X₁-X₂ > 4) = 0.66276

Therefore, the probability that X₁-X₂ exceeds 4 is 0.66

(b) The probability that 4.8 ≤ X₁-X₂   ≤ 5.6.

P(4.8 < X₁-X₂ < 5.6) = P((4.8-​​​​​​​ 6) /4.72738 < ((X₁-X₂ )-​​​​​​​ ) / ()  < (5.6-​​​​​​​ 6) /4.72738 )

P(4.8 < X₁-X₂ < 5.6) = P((4.8-​​​​​​​ 6) /4.72738 < Z  < (5.6-​​​​​​​ 6) /4.72738 )

P(4.8 < X₁-X₂ < 5.6) = P( -0.25 < Z  < -0.08 )

P(4.8 < X₁-X₂ < 5.6) = P(Z  < -0.08 ) - P(Z  < -0.25 ) (form the normal distribution)

P(4.8 < X₁-X₂ < 5.6) = 0.46812 - 0.40129

P(4.8 < X₁-X₂ < 5.6) = 0.06683