The weights of a certain variety of Giant pumpkins are normally distributed as f

Question

The weights of a certain variety of Giant pumpkins are normally distributed

as follows: µ = 300 pounds, s = 20 pounds. Fill in the blanks. DRAW A NORMAL CURVE INDICATING µ AND ± 3s ABOVE THIS PROBLEM, THEN ANSWER THE FOLLOWING.

a) _________________What percent of these pumpkins weigh between 280 and 320 pounds?

b) _________________What percent of these pumpkins weigh between 300 and 320 pounds?

c)   _________________What percent of these pumpkins weigh less than 360 pounds?

d) _________________What percent of these pumpkins weigh more than 360 pounds?

e) ___________________What weight corresponds to a z-score of -1.5?

f)   ________________How many standard deviations away from the mean is a 272 pound pumpkin?

____________________What percent of pumpkins weigh less than 272 pounds?

h) ____________________What percent of pumpkins weigh more than 272 pounds?

i) ____________________What percent of pumpkins weigh between 272 and 312 pounds?

Explanation / Answer

SOlving first 4 subparts

1) 320 is 1 standard deviation away from mean in +ve direction,while 280 in negative direction. Hence ,we have to find the probability between z= 1 and z=-1.

Hence,probability = 1- 2 [P(z<1)]= 1-2*0.158655=0.6827

2)Probability = 1- P(z>1) = 1-0.158655=0.841345

3) z = [360-300]/20 = 3

Probability = P(z<3 ) = 1- P(z>3) = 1- 0.001349= 0.99865

4) Probability = 1- Probability in part 3 = 1-0.99865=0.001349

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