Question

Suppose the heights of 18-year-old men are approximately normally distributed, with mean 69 inches and standard deviation 4 inches.

(a) What is the probability that an 18-year-old man selected at random is between 68 and 70 inches tall? (Round your answer to four decimal places.)

(b) If a random sample of seventeen 18-year-old men is selected, what is the probability that the mean height x is between 68 and 70 inches? (Round your answer to four decimal places.)

(c) Compare your answers to parts (a) and (b). Is the probability in part (b) much higher? Why would you expect this?

The probability in part (b) is much higher because the standard deviation is smaller for the x distribution.

The probability in part (b) is much higher because the mean is larger for the x distribution.    

The probability in part (b) is much higher because the mean is smaller for the x distribution.

The probability in part (b) is much lower because the standard deviation is smaller for the x distribution.

The probability in part (b) is much higher because the standard deviation is larger for the x distribution.

The method of tree ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population of x values has an approximately normal distribution.

1264 1285 1313 1187 1268 1316 1275 1317 1275

(a) Use a calculator with mean and standard deviation keys to find the sample mean year x and sample standard deviation s. (Round your answers to the nearest whole number.)

x =

s =

(b) Find a 90% confidence interval for the mean of all tree ring dates from this archaeological site. (Round your answers to the nearest whole number.)

lower limit  

upper limit  

Answer 1

Solution :

Given that mean μ = 69 and standard deviation σ = 4

(a)
=> P(68 < x < 70) = P((68 - 69)/4 < (x - μ)/σ < (70 - 69)/4)

= P(-0.25 < Z < 0.25)

= P(Z < 0.25) − P(Z < −0.25)

= 0.5987 - 0.4013

= 0.1974

(b)
=> P(68 < x < 70) = P((68 - 69)/(4/sqrt(17) < (x - μ)/(σ/sqrt(n)) < (70 - 69)/(4/sqrt(17)))

= P(-1.0308 < Z < 1.0308)

= P(Z < 1.0307) − P(Z < −1.0307)

= 0.8485 - 0.1515

= 0.6970

(c)
=> The probability in part (b) is much higher because the standard deviation is smaller for the x distribution.

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Solution :

Given that 1264 1285 1313 1187 1268 1316 1275 1317 1275
  
(a)
=> mean x = sum of terms/number of terms

= 11500/9

= 1277.7778

= 1278 (nearest whole number)

=> standard deviation s = 40 (nearest whole number)

(b)
=> df = n - 1 = 8

=> for 90% confidence interval, t = 1.860

=> A 90% confidence interval of the mean is

=> x +/- t*s/sqrt(n)

=> 1278 +/- 1.860*40/sqrt(9)

=> (1253.2,1302.8)

=> (1253,1303) (nearest whole number)

=> lower limit = 1253

=> upper limit = 1303