question 10 of 10 in the united states, egg sizes are set by the department of agriculture. a “large” egg…

question 10 of 10 in the united states, egg sizes are set by the department of agriculture. a “large” egg, for example, weighs between 57 and 64 grams. suppose the weights of eggs produced by hens owned by a particular farmer are approximately normally distributed with a mean of 55.8 grams and a standard deviation of 7.5 grams. (a) what proportion of these eggs weigh less than 50 grams? (round to 4 decimal places) (b) what percent of these eggs would be classified as “large”? % (round to 2 decimal places)

question 10 of 10 in the united states, egg sizes are set by the department of agriculture. a “large” egg, for example, weighs between 57 and 64 grams. suppose the weights of eggs produced by hens owned by a particular farmer are approximately normally distributed with a mean of 55.8 grams and a standard deviation of 7.5 grams. (a) what proportion of these eggs weigh less than 50 grams? (round to 4 decimal places) (b) what percent of these eggs would be classified as “large”? % (round to 2 decimal places)

Answer

Explanation:

Step1: Calculate z - score for part (a)

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 50$, $\mu=55.8$, and $\sigma = 7.5$. So $z=\frac{50 - 55.8}{7.5}=\frac{-5.8}{7.5}\approx - 0.77$.

Step2: Find proportion for part (a)

Using the standard normal distribution table, $P(Z < - 0.77)=0.2206$.

Step3: Calculate z - scores for part (b)

For $x_1 = 57$, $z_1=\frac{57 - 55.8}{7.5}=\frac{1.2}{7.5}=0.16$. For $x_2 = 64$, $z_2=\frac{64 - 55.8}{7.5}=\frac{8.2}{7.5}\approx1.09$.

Step4: Find proportion for part (b)

$P(0.16<Z<1.09)=P(Z < 1.09)-P(Z < 0.16)$. From the standard - normal table, $P(Z < 1.09)=0.8621$ and $P(Z < 0.16)=0.5636$. So $P(0.16 < Z < 1.09)=0.8621 - 0.5636=0.2985$. To convert to percentage, we multiply by 100, getting $29.85%$.

Answer:

(a) $0.2206$ (b) $29.85%$