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The demand for walnut pastries at Joanne’s Bakery on Saturday mornings is N (30,

ID: 3055099 • Letter: T

Question

The demand for walnut pastries at Joanne’s Bakery on Saturday mornings is N(30, 8). Joanne currently makes 36 walnut pastries for Saturday mornings. If four Saturday mornings are randomly selected, what is the probability that Joanne will have run out of walnut pastries on every one of the Saturdays?

A) 0.0026

B) 0.2266

C) 0.9064

D) 0.0668

A market research study about brand preference for a particular product among males and females revealed the following: 70% of consumers of this product are men. 40% of consumers prefer Brand A 50% of consumers prefer Brand B 10% of consumers prefer other brands. Half of all males prefer Brand A Two-thirds of all females prefer Brand B What is the probability that a consumer randomly chosen from among those who prefer Brand A will be a female? A) 0.050 B) 0.125 C) 0.167 D) 0.500 A market research study about brand preference for a particular product among males and females revealed the following: 70% of consumers of this product are men. 40% of consumers prefer Brand A 50% of consumers prefer Brand B 10% of consumers prefer other brands. Half of all males prefer Brand A Two-thirds of all females prefer Brand E From this we know that brand preference and gender are: A) independent because P(Prefer Brand A | Male)-P(Prefer Brand A and Male) B) not independent because P(Gender | Brand Preferred) -P(Gender) for all brand, gender combinations. C) independent because more males prefer Brand A and more females prefer Brand B D) not independent because P(Prefer Brand A | Female) f P(Prefer Brand A)

Explanation / Answer

1)P(Brand A and female) =P(Brand A)-P(brand A and male) =0.4-0.7*0.5=0.05

therefore P(female|brand A) =P(Brand A and female)/P(brand A) =0.05/0.4 =0.125

option B is correct

2)option D is correct ; not independent,........

3)

hence probability of  run out of walnut pastries on every one of the Saturdays =(0.2266)4 =0.0026

option A is correct

for normal distribution z score =(X-?)/? here mean=       ?= 30 std deviation   =?= 8
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