Question

Assume that the distribution of squirrel weights are normally distributed with mean=1.7 lbs and standard dev.= 0.5 lbs. Suppose we want to find the proportion of squirrels that weigh between 0.8 and 1.2 lbs.

Suppose we want to find the proportion of squirrels that weigh between 0.8 and 1.2 lbs.

Standardize the weights of 0.8 lbs and 1.2 lbs. Write the z-scores corresponding to the two weights below.

Answer 1

X : Squirrel weights

X normally distributed with mean=1.7 lbs and standard dev.= 0.5 lbs

Proportion of squirrels that weigh between 0.8 and 1.2 lbs i.e Probability of a squirrel that weigh between 0.8 and 1.2 lbs = P(0.8X1.2)

P(0.8X1.2) = P(X1.2) - P(X0.8)

P(X1.2);

Z-score = (X-mean)/Standard deviation

Z-score for 1.2 = (1.2 - 1.7) / 0.5 = -0.5/0.5 = - 1

Z-score for 1.2 = -1

P(X1.2) = P(Z-1)

From Standard normal tables;

P(Z-1) = 0.1587

P(X1.2) = P(Z-1) = 0.1587

P(X0.8);

Z-score = (X-mean)/Standard deviation

Z-score for 0.8 = (0.8 - 1.7) / 0.5 = -0.9/0.5 = - 1.8

Z-score for 0.8 = -1.8

P(X0.8) = P(Z-1.8)

From Standard normal tables;

P(Z-1.8) = 0.0359

P(X0.8) = P(Z-1.8) = 0.0359

P(0.8X1.2) = P(X1.2) - P(X0.8) = 0.1587-0.0359=0.1228

P(0.8X1.2) = 0.1228

Proportion of squirrels that weigh between 0.8 and 1.2 lbs i.e Probability of a squirrel that weigh between 0.8 and 1.2 lbs = P(0.8X1.2) = 0.1228

Proportion of squirrels that weigh between 0.8 and 1.2 lbs = 0.1228

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Z-score for 1.2 = -1

Z-score for 0.8 = -1.8