Question

weights of the pacific yellowfin tuna follow a normal distribution with mean weight 68lbs and standard deviation 12lbs. For a randomly caught Pacific yellowfin tuna, what is the probability that the weight is

a) less than 50 lbs?

b) more than 80 lbs?

c) between 50lbs and 80 lbs?

Answer 1

Solution :

Given that ,

mean = = 68

standard deviation = = 12

a)

P(x < 50) = P[(x - ) / < (50 - 68) / 12]

= P(z < -1.5)

= 0.0668

Probability = 0.0668

b)

P(x > 80) = 1 - P(x < 80)

= 1 - P[(x - ) / < (80 - 68) / 12)

= 1 - P(z < 1)

= 1 - 0.8413

= 0.1587

Probability = 0.1587

c)

P(50 < x < 80) = P[(50 - 68)/ 12) < (x - ) /  < (80 - 68) / 12) ]

= P(-1.5 < z < 1)

= P(z < 1) - P(z < -1.5)

= 0.8413 - 0.0668

= 0.7745

Probability = 0.7745