Question

Q3.(15) It is known that the weight of pumpkin in one farm follows normal distribution with average weight 25 lb, standard deviation of 9 lb.
1.(5) Find the probability that the weight of one randomly selected pumpkin is larger than 40lb.
2.(5) Find the probability that the weight of one randomly selected pumpkin is between 20 lb and 30 lb, inclusive.
3.(5) If we have a random sample of 36 pumpkins, find the probability that the mean weight is between 20 lb and 30 lb, inclusive.

Answer 1

Solution :

Given that ,

1) P(x > 40) = 1 - p( x< 40)

=1- p P[(x - ) / < (40 - 25) / 9]

=1- P(z < 1.67)

= 1 - 0.9525

= 0.0475

2) P(20 < x < 30) = P[(20 - 25)/ 9) < (x - ) /  < (30 - 25) / 9) ]

= P(-0.56 < z < 0.56)

= P(z < 0.56) - P(z < -0.56)

Using z table,

= 0.7123 - 0.2877

= 0.4246

3) n = 36

= 25

= / n = 9 / 36 = 1.5

P(20 < < 30)  

= P[(20 - 25) / 1.5 < ( - ) / < (30 - 25) / 1.5)]

= P( -3.33 < Z < 3.33)

= P(Z < 3.33) - P(Z < -3.33)

Using z table,  

= 0.9996 - 0.0004

= 0.9992