3. The mean amount purchased by a typical customer at Wal Grocery Store is $23.5

Question

3. The mean amount purchased by a typical customer at Wal Grocery Store is $23.50 with a standard deviation of S5.00 Assume the distribution of amounts purchased follows the normal distribution. For a sample of 4g customers answer the following questions a. What is the likelihood the sample mean is at least $24.207 P2.20 b. What is the likelihood the sample mean is greater than $24.20 but $25.502 24.20 442 9 . CE) c. Within what limits will 94% of the sample means occur. PG+2%20-1-r7ZM -2-241 10-28.59 4. a) The Accounting Department has six faculty, three of whom are female. In a gundog sampleof faculty, what is the probability that none is female? -10 Ga or6.5 b) T-Printing Inc. purchases plastic cups on which to print logos for sporting events, proms, and other special occasions. The owner received alarge shipment this morning. He found 1% to be defective. For a random sample of 200 cups, find the probability that two or more of the cups ane defective

Explanation / Answer

Ans:

3)standard error of mean=5/sqrt(49)=5/7=0.714

a)

z=(24.20-23.5)/0.714=0.98

P(z>=0.98)=0.1635

b)

z=(25.5-23.5)/0.714=2.8

P(0.98<=z<=2.8)=P(z<=2.8)-P(z<=0.98)=0.9974-0.8365=0.1610

c)z value or 94 CI% 1.88

94% CI

=23.5+/-1.88*0.714

=23.5+/-1.34

=(22.16,24.84)

4)a)Probability of being a female=3/6=0.5

n=3

Binomial distribution with n=3,p=0.5

P(none is female)=3C0*0.50*(1-0.5)3=0.53=0.125

b)Binomial distribution with n=200,p=0.01

P(2 or more)=1-P(less than 2)=1-P(0 defective)-P(1 defective)=1-200C0*0.010*0.99200-200C1*0.011*0.99199

=1-0.1340-0.2707

=0.5953

or use excel function:

P(x>=2)=1-P(x<=1)=1-BINOMDIST(1,200,0.01,TRUE)=0.5953

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