Question

50 pure bread Maltese were analyzed and the results were that the average Maltese weighs 7.7 lbs with a variance of 1.25lbs. However, the Maltese Kennel Club needs some more information...

a. the award winning Maltese Princess weighs 10.3lbs, what percentile and above is she in?

b. 3rd place winner, Spot, weighs only 6 lbs, what proportion or less is he in?

c. What is the portion of dogs weight between Princess and Spot

d. What is the weight of the dogs in the top 14 percent?

e. How many dogs weigh between 9 lbs and 10 lbs?

Answer 1

Let X denote the weight of dog

Mean, = 7.7 lbs

Standard deviation, = 1.25 lbs

(a) The required percentile = P(X < 10.3)

= P{Z < (10.3 - 7.7)/1.25} = P(Z < 2.08)

= 0.9812 = 98.12% percentile

(b) The required proportion = P(X < 6)

= P{Z < (6 - 7.7)/1.25}

= P(Z < -1.36)

= 0.0869

(c) The portion of dogs between Princess and Spot = P(6 < X < 10.3)

= P(-1.36 < Z < 2.08) = 0.8943

(d) Let the required weight be x

-> P(X > x) = 0.14

Corresponding z value = 1

Thus, x = 7.7 + 1.25 = 8.95 lbs

(e) Probability that a randomly selected dog weigh between 9 lbs ans 10 lbs

= P(9 < X < 10)

= P(1.04 < Z < 1.84)

= 0.1163

Thus , number of dogs = 50*0.1163 = 5.815

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