The hotel room rate in New York City is normally distributed with a mean of $20s

Question

The hotel room rate in New York City is normally distributed with a mean of $20s per night. Assume that the standard deviation is unknown. (See exercise 25 on page 289 of your textbook for a similar problem.) stion 4 If 23% of the New York City hotel room rates are more than S222per night, what is the varane7(Remember the label) dollars 2 9A2 minfes minutes 2 28% of consumers read the ingredients listed on a product's label. (See exercise 35 on page 332 of your textbook for a similar problem 57 hours*2 poundsA2 points 2 Question 5 For a sample of 230 consumers, what is the probability that between 51 and 74 of them read the ingredients isted on a product's labe? 60 62 63 64 The Economic Policy Institute reports that the average entry-level wage for male college graduates is $22.01 per hour and for female college graduates is $18.52 per hour. The 65 standard deviation for male graduates is $3.80 and for female graduates is $2.83. Assume wages are normally distributed. (See exercise 23 on page 326 of your textbook for a 66 similar problem.) 67 Question 46 male eraduates are chosen, find the probability the sample average entry-level wage is at least $21.41 70

Explanation / Answer

#4.
mean = 205

Using standard z table, z-value for 0.23 is 0.7388

Now using central limit theorem,

z = (x - mu)/sigma
sigma = (x - mu)/z = (222 - 205)/0.7388 = 23.0103

#5.
p = 0.28, n = 230
mean = np = 230 * 0.28 = 64.4
std. dev. = sqrt(np(1-p)) = sqrt(230*0.28*0.72) = 6.8094

using central limit theorem
P(51 < X < 74)
= P((51 - 64.4)/6.8094 < z < (74 - 64.4)/6.8094)
= P(-1.9679 < z < 1.4098)
= P(z < 1.4098) - P(z < -1.9679)
= 0.9207 - 0.0245 (values are taken from standard z table)
= 0.8962

#6.
mean = 22.01
std. dev. = 3.8

P(X >= 21.41)
= P(z >= (21.41 - 22.01)/(3.8/sqrt(46)))
= P(z >= -1.0709)
= 1 - P(z < -1.0709)
= 1 - 0.1421 (values are taken from standard z table)
= 0.8579

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