Question

Suppose the heights of 18-year-old men are approximately normally distributed, with mean 67 inches and standard deviation 3 inches.

b) If a random sample of thirteen 18-year-old men is selected, what is the probability that the mean height x is between 66 and 68 inches? (Round your answer to four decimal places.)

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Let x represent the dollar amount spent on supermarket impulse buying in a 10-minute (unplanned) shopping interval. Based on a certain article, the mean of the x distribution is about $22 and the estimated standard deviation is about $9.

(b) What is the probability that x is between $19 and $25? (Round your answer to four decimal places.)

(c) Let us assume that x has a distribution that is approximately normal. What is the probability that x is between $19 and $25? (Round your answer to four decimal places.)

Answer 1

Solution :

Given that,

mean = = 67

standard deviation = = 3

n = 13

b)

= 67

= / n = 3 / 13 = 0.8321

P( 66 < < 68) = P((66 - 67) / 0.8321) <( - ) / < (68 - 67) / 0.8321))

= P(-1.2018 < Z < 1.2018)

= P(Z < 1.2018) - P(Z < -1.2018) Using z table,

= 0.8853 - 0.1147

= 0.7706

Probability = 0.7706

mean = = $22

standard deviation = = $9

b)

P($19 < x < $25) = P((19 - 22)/ 9) < (x - ) / < (25 - 22) / 9) )

= P(-0.33 < z < 0.33)

= P(z < 0.33) - P(z < -0.33)

= 0.6293 - 0.3707

= 0.2586

Probability = 0.2586