Use the following information to answer questions 4, 5, 6 and 7. The weights of

Question

Use the following information to answer questions 4, 5, 6 and 7. The weights of adult Fox Terriers in US (a dog breed) are normally distributed, with a mean of 15 pounds and a standard deviation of 3 pounds. The probability that a randomly chosen Fox Terrier weighs between 9 pounds and 18 pounds is closest to (a) 0.7961 (b) 0.2514 (c) 0.8185 (d) 0.0475 (e) .9725 4. 5. To be in the top 13% of the weights, a Fox Terrier should weigh at least (a) 18.39 pounds. (b) 25.27 pounds. (c) 17.52 pounds (d) 20.64 pounds (e) 15 pounds. 6. A random sample of 100 Fox Terriers is drawn from this population. Identify the mean , and standard error ??. Of the sample mean weight X. 7. Find the probability that the sample mean weight x exceeds 16 pounds. (a) .9996 (b) .5000 (c) .6293 (d) .3707 (e) .0004

Explanation / Answer

Given: Mean,u = 15

Standard Deviation, s = 3

Let X denote the weight of the fox Terrier

(4) To find, P(9< X < 18)

For X = 9, Z = (9 - 15)/3 = -2

For X = 18, Z =(18 - 15)/3 = 1

Thus, P(9< X < 18) = P(-2< Z <1) = 0.8185(option c)

(5) P(X > xo) = 0.13

To find xo.

from Z table, P(Z > 1.13) = 0.13

Thus, (xo - 15)/3 = 1.13

-> xo = 18.39 pounds (option a)

(6) Mean of the sample mean weight remains same as the population mean weight = 15

Standard error of the sample mean weight = s/sqrt(n) = 3/sqrt(100) = 0.3

Thus, the correct answer is option (a)

(7) P(sample mean weight > 16)

Z value corresponding to sample mean weight = (16 - 15)/standard error = 1/0.3 = 3.333

Thus, the required probability = P(Z > 3.333) = .0004(option c)

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---