Question

Suppose a research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week. A sample of 100 steady smokers revealed that the sample mean is $20. The population standard deviation is $5. What is the probability that a sample of 100 steady smokers spend between $19 and $21?

Group of answer choices

1.0000

0.0228

0.4772

0.9544

Answer 1

Solution :

Given that,

mean = = $20

standard deviation = = $5

= / n = 5 / 100 = 0.5

= P[(19 - 20) / 0.5< ( - ) / < (21 - 20) / 0.5)]

= P(-2 < Z < 2)

= P(Z < 2) - P(Z < -2)

= 0.9772 - 0.0228

= 0.9544

Probability = 0.9544