Question

28% of consumers read the ingredients listed on a product's label.

Answer 1

Answer:

The probability that between 64 and 90 of them read the ingredients listed on a product's label is 0.82.

Step-by-step explanation:

We know the proportion of the population, that is π=0.28.

The standard deviation for this sampling distribution is

$\sigma=\sqrt{\frac{\pi(1-\pi)}{N} } =\sqrt{\frac{0.28*0.72}{256} }= 0.028$

The value 64 represents a proportion of p=64/256=0.25, and the value 90 represents p=90/256=0.35.

With these values, we can calculate the z-values.

$z_1=\frac{p_1-\pi}{\sigma} =\frac{0.25-0.28}{0.028} =\frac{-0.03}{0.028} =-1.07\\\\z_2=\frac{p_2-\pi}{\sigma} =\frac{0.35-0.28}{0.028} =\frac{0.07}{0.028} =2.5$

The probability that between 64 and 90 read the ingredients is:

$P=P(64$

The probability that between 64 and 90 of them read the ingredients listed on a product's label is 0.82.