The diameter of a brand of tennis balls is approximately normally distributed, w

Question

The diameter of a brand of tennis balls is approximately normally distributed, with a mean of 2.51 inches and a standard deviation of 0.03 inch. A random sample of 11 tennis balls is selected.

a. What is the probability that the sample mean is less than 2.49 inches?

P(X < 2.49)=.0136

b.

What is the probability that the sample mean is between 2.50 and 2.52 ?inches?

P(2.50 < X < 2.52)

d. The probability is 57 % that the sample mean will be between what two values symmetrically distributed around the population mean?

The lower bound is __ inches. The upper bound is __ inches.

Explanation / Answer

Solution : Given that mean ? = 2.51 , standard deviation ? = 0.03 , n = 11

a. P(x < 2.49) = P((x - ?)/(?/sqrt(n)) < (2.49 - 2.51)/(0.03/sqrt(11)))

= P(Z < -2.2111)

= 0.0136

b. P(2.50 < X < 2.52) = P((x - ?)/(?/sqrt(n)) < Z < (x - ?)/(?/sqrt(n)))

= P((2.50 - 2.51)/(0.03/sqrt(11)) < Z < (2.52 - 2.51)/(0.03/sqrt(11)))

= P(-1.1055 < Z < 1.1055)

= 0.7330

d. P(-z < Z < z) = 0.57
P(Z < z) - P(Z < 0) = 0.285
P(Z < z) = 0.285 + 0.5
P(Z < z) = 0.785

=> z = 0.7892

lower bound = ? - Z*?/sqrt(11)
= 2.51 - 0.7892*0.03/sqrt(11)
= 2.5029

Upper bound = ? + Z*?/sqrt(11)
= 2.51 + 0.7892*0.03/sqrt(11)
= 2.5171

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.