Question

Probability-1  

A. A bag of potato chips is labelled as weighing 255 grams. However, in fact, the weights of bags of potato chips from this manufacturer are normal distributed with mean 255 grams and standard deviation 5 grams.

a. What is the probability that a randomly selected bag of potato chips will weigh less than 248 grams?

b. What percentage of bags of potato chips weigh between 253 and 257 grams?

c. What is the probability that a randomly selected bag of potato chips will weigh 255 grams?

d. Determine the value such that 95% of bags of potato chips have weight greater than.

B. Thyroid stimulating hormone (TSH) in the population of individuals without thyroid disorders has a density with mean 2 mIU/L and standard deviation 0.64 mIU/L.

a. Do you have enough information to compute (at least approximately) the proportion of individuals in this population who have TSH greater than 3 mIU/L? If so, provide this value. If not, explain why.

b. Do you have enough information to compute (at least approximately) the probability that the average TSH in a random sample of 100 individuals from this population will be greater than 2.1 mIU/L? If so, provide this value and a justification. If not, explain why.

Answer 1

Question 1:

The distribution given here is:

a) The required probability here is computed as:

P(X < 248 )

Converting this to a standard normal variable, we get:

Getting this from the standard normal tables, we get:

Therefore 0.0808 is the required probability here.

b) The required probability here is computed as:

P( 253 < X < 257 )

Converting this to a standard normal variable, we get:

Getting this from the standard normal tables, we get:

Therefore 31.08% is the required percentage here.

c) Now the probability that a randomly selected bag of potato chips will weigh 255 grams is computed as:

Therefore 0.0798 is the required probability here.

d) Let the required value for which 95% bag weigh higher than be K. Then, we have here:

P(X > K ) = 0.95

Converting this to a standard normal variable, we get:

From standard normal tables, we get:

P(Z > -1.645 ) = 0.95

Therefore, we get here: