Use the normal distribution of fish lengths for which the mean is 11 inches and

Question

Use the normal distribution of fish lengths for which the mean is 11 inches and the standard deviation is 4 inches. Assume the variable x is normally distributed. (a) What percent of the fish are longer than 13 inches? (b) If 500 fish are randomly selected, about how many would you expect to be shorter than 8 inches? (a) Approximately offish are longer than 13 inches. (Round to two decimal places as needed.) (b) You would expect approximately fish to be shorter than 8 inches. (Round to the nearest fish.)

Explanation / Answer

(a)

Mean = 11 inches

Std deviation = 4 inches

Proportion of dish longer than 13 inches is,

P(X > 13) = P ( Z > (13-11)/4) = P (Z > 0.5)

By z-table, P(Z > 0.5) = 0.3085

So, Approximately, 30.85% of fish are longer than 13 inches.

(b)

Proportion of dish shorter than 8 inches is,

P(X < 8) = P ( Z < (8-11)/4) = P (Z < -0.75)

By z-table, P(Z < - 0.75) = 0.2266

Number of fish shorter than 8 inches = 500 * 0.2266 = 113.3

You would expect approximately 113 fish to be shorter than 8 inches.

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