A chocolate bar machine makes bars which average 10 ounces with an SD of 1.5 oun

Question

A chocolate bar machine makes bars which average 10 ounces with an SD of 1.5 ounces. Assume that the amount would be normally distributed. A. What is the probability that a randomly selected bar has more than 8oz of chocolate? (enter a, b, c, d or e) a. 0.25 b. 0.09 c. 0.75 d. 0.91 e. 0.08 B. What is the probability that a randomly selected bar has between 11 to 12oz of chocolate? (enter a, b, c, d or e) a. 0.16 b. 0.66 c. 0.75 d. 0.95 e. 0.84 C. How many of the next 100 bars will weight less than 12.8 ounces? (enter a, b, c, d or e) a. 3 b. 43 c. 57 d. 96 e. 93 'Lawsuit Claims Nightclub Fired for Being Too Short was a headline of Foxs News New York. According to the report, two women filed suit against their previous employer, a prominent New York nightclub, after the two aspiring actresses, 3in and 4in respectively, were find for being too short to meet the clubs image requirement. The report also claims that "are both roughly the same size as the average American woman at 5 3in tall" Suppose that the height of a randomly selected adult woman is approximately normal with mean 63in. (5 3in) and standard deviation 2in. The waitresses claim that this nightclub's height policy excludes 90% of all women from being waitresses there. Find the minimum height requirement that would exclude 90% of all women from being waitresses at this Round your answer to the nearest inch. Customers at the local Subway soon order an ice with their meal 65% of the time. Use the Binomial distribution to find the answer to the following A. Find the probability that among the next 8 customers, at least 3 will order ice with their meal. (enter options a, b, c or d. Do not enter a number.) a. 0.0808

Explanation / Answer

6. a From information given, mu=10, sigma=1.5, since the amount of chocolate in the bar is normally distributed, use z score formula [z=(X-mu)/sigma, where, X is raw score, mu is population mean, and sigma is population standard deviation] to compute the required probability.

P(X>8)=P[Z>(8-10)/1.5]=P(Z>-1.33)

=1-0.0918 (the z table gives area under standard normal curve to the left of z, thus, subtract the area from 1)

=0.9082~0.91 (d)

b. Compute z scores for two raw scores, X1=11 and X2=12.

z1=(11-10)/1.5=0.67 and z2=(12-10)/1.5=1.33

Both the z scores are above mean, therefore, find areas between z1 and mean as well as z2 and mean. Subtract the smaller area from the larger area. The area between 0.67 and mean is 0.2486 and area between 1.33 and mean is 0.4082.

Thus, P(11<X<12)=0.4082-0.2486=0.1596 ~0.16 (a)

c. P(X<12.8)=P[Z<(12.8-10)/1.5]=P(Z<1.87)=0.9693~0.97 (d)

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