g Eq Score: 0 of 4 pts 6.2.7-T 3 of 11 (9 complete) HW Score: 76.58%, : Ques The

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g Eq Score: 0 of 4 pts 6.2.7-T 3 of 11 (9 complete) HW Score: 76.58%, : Ques The annual ground coffee expenditures for households are approximately normally distributed with a mean of $48.51 and a standard deviation of $13.00 atic a. Find the probability that a household spent less than $25.00. b. Find the probability that a household spent more than $60.00. c. What proportion of the households spent between $30.00 and $40.00? d, 97.5% of the households spent less than what amount? Geoc a. The probability that a household spent less than $25.00 is Round to four decimal places as needed.) ex N ns s an us C Enter your answer in the answer box and then click Check Answer. Clear All tic Geometry

Explanation / Answer

Solution:

a) mean = 48.51
Std Dev (s) = 13
To find : P( x < 25 )
z = ( x - )/s = ( 25 - 48.51 )/13 = -1.8085

P( x < 25) = P( Z < -1.8084)
= 0.0351 _ _ _ _ _ _ _ _ _ (From Z-table)
= 3.51%
b) Find : P( x > 60 )
z = ( x - )/s = ( 60 - 48.51 )/13 = 0.8838

P( x > 60) = P( Z > 0.8838) = P( Z < 0.1894)
= 0.1894 _ _ _ _ _ _ _ _ _ (From Z-table)
= 18.94%

c) Find : P( 30 < X < 40 )
We convert to standard normal form, by z = (x - )/s
so z1 = (30 - 48.51 )/13 = -1.4238
& z2 = (40 - 48.51 )/13 = -0.6546

P( 30 < X < 40) = P(z1 < Z < z2) = P( Z < -1.42) - P(Z < -0.65)
= 0.2578 - 0.0778
= 0.18 _ _ _ _ _ _ _ _ _ (From Z-table)

d) 97.5th percentile is given by:
0.975 = P(x < x0) = P( (x-)/s < (x0 -)/s)

from z table, we get (x0-)/s = 1.9599

This gives x0 = 73.9887

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