Question

1, 33% of consumers read the ingredients listed on a product's label. For a sample of 265 consumers, what is the probability that between 80 and 103 of them read the ingredients listed on a product's label?

2.

The hotel room rate in New York City is normally distributed with a mean of $294 per night. Assume that the standard deviation is unknown.

If 23% of the New York City hotel room rates are more than $328 per night, what is the variance? (Remember the label.)

3.

The Economic Policy Institute reports that the average entry-level wage for male college graduates is $22.23 per hour and for female college graduates is $18.97 per hour. The standard deviation for male graduates is $3.94 and for female graduates is $3.02. Assume wages are normally distributed.

If 30 female graduates are chosen, find the probability the sample average entry-level wage is at least $19.85

Answer 1

1.

Sample proportion, p = 0.33

Standard error of sample proportion, SE = = 0.0289

Probability that between 80 and 103 of them read the ingredients listed on a product's label

= P(80/265 < p < 103/265)

= P(0.3019 < p < 0.3887)

= P(p < 0.3887) - P(p < 0.3019)

= P[Z < (0.3887 - 0.33) / 0.0289] - P[Z < (0.3019 - 0.33) / 0.0289]

= P[Z < 2.03] - P[Z < -0.97]

= 0.9788 - 0.1660

= 0.8128

2.

Let X be the hotel room rate in New York City . Then,

P(X > 328) = 0.23

P[Z > (328 -  294) / ] = 0.23

(328 -  294) / = 0.7388

= 34 / 0.7388 = 46.02

Variance = 46.02² = 2117.84

3.

Standard error of sample mean = 3.02 / = 0.551374

Probability the sample average entry-level wage is at least $19.85 = P( 19.85)

= P(Z (19.85 - 18.97)/0.551374]

= P[Z 1.60]

= 0.0548