Question

It is known that the weight of a certain breed of dog has a normal distribution with a mean of 25.65 pounds and a standard deviation of 4.87 pounds.

1) What is the probability that an individual dog of this breed will weigh between 20 and 30 pounds?

2) What is the probability that an individual dog of this breed will weigh more than 25 pounds?

3) What is the probability that a litter of 4 puppies of this breed will grow up to have an average weight of less than 20 pounds?

4) A dog of this breed is classified as “overweight” if its weight is in the top 15%. How much would such a dog have to weigh at minimum to be classified thus?

5) A dog of this breed is classified as “underweight” if its weight is in the bottom 10%. How much would such a dog have to weigh at maximum to be classified thus?

Answer 1

Solution :

Given that ,

mean = = 25.65

standard deviation = = 4.87

1)P(20 < x < 30) = P[(20 - 25.65)/4.87 ) < (x - ) /  < (30 - 25.65) /4.87 ) ]

= P(-1.16 < z < 0.89)

= P(z < 0.89) - P(z < -1.16)

= 0.8133 - 0.123 = 0.6903

Probability = 0.6903

2) P(x > 25) = 1 - P(x < 25)

= 1 - P[(x - ) / < (25 - 25.65) /4.87 )

= 1 - P(z < -0.13)

= 1 - 0.4483 = 0.5517

Probability = 0.5517

3) n = 4

= / n = 4.87 / 4 = 2.435

P( < 20) = P(( - ) / < (20 - 25.65) / 2.435)   

P(z < -2.32)

0.0102

Probability = 0.0102

4) Using standard normal table ,

P(Z > z) = 15%

1 - P(Z < z) = 0.15

P(Z < z) = 1 - 0.15 = 0.85

P(Z < 1.04) = 0.85

z = 1.04

Using z-score formula,

x = z * +

x = 1.04 * 4.87 + 25.65 = 30.71

would such a dog have to weigh at minimum to be classified Is 30.71

5) Using standard normal table ,

P(Z < z) = 10%

P(Z < z) = 0.1

P(Z < -1.28) = 0.1

z = -1.28

Using z-score formula,

x = z * +

x = -1.28 * 4.87 + 25.65 = 19.42

would such a dog have to weigh at maximum to be classified is 19.42