Question

Question 21 of 24 (1 point) 6.3 Section Exercise 8 A survey found that the American family generates an average of 17.2 pounds of glass garbage each year. Assume the standard deviation of the distribution is 2.5 pounds. Find the probability that the mean of a sample of 58 families will be between 17.1 and 18.1 pounds. Assume that the sample is taken from a large population and the correction factor can be ignored. Round your final answer to four decimal places and intermediatez-value calculations to two decimal places. Source: Michael D. Shook and Robert L. Shook, The Book of Odds. P(17.1 < X < 18.1)=C

Answer 1

Solution :

Given that,

mean = $\mu$ = 17.2

standard deviation = $\sigma$ = 2.5

$\sigma$$\bar x$ = $\sigma$ / $\sqrt$ n = 2.5 / $\sqrt$ 58 = 0.3283

= P[(17.1 - 17.2) / 0.3283< ($\bar x$ - $\mu$ $\bar x$) / $\sigma$ $\bar x$ < (18.1 - 17.2) / 0.3283)]

= P(-0.30 < Z < 2.74)

= P(Z < 2.74) - P(Z < -0.30)

= 0.9969 - 0.3821

= 0.6148

P(17.1 < $\bar x$ < 18.1) = 0.6148