Answer: a) 79.10 % b) 7% c)19.85 %
Step-by-step explanation:
a) Z = ( × - μ ) ÷ σ Where μ mean of population σ standard deviation Z is the abscissa to give the area or probability we are looking for associated to the value 10 ounces ( × )
So: Z = ( 10 - 9 ) ÷ 1.2 ⇒ Z = 1/1.2 ⇒ Z = 0.83 it has to be below this value
From Z tables we get P [ Z ≤ 0.82 ] = 0.7910 0r 79.10 %
b) Following the same procedure: We look for
P [ Z > ( x - μ ) ÷ σ ] ⇒ Z = ( 12 -10 ) ÷ 1.2 = 2.5
From Z table we get the area under the curve from the left tail up to the point Z < 2.5 ( 2.5 not included) but we were asked for the area out of that previous so 1- 0.9930 = 0.007 is the area we are looking for
So P (b) = 0.007 or 7 %
Finally between the two points above mentioned ( 10 ≤ Z ≤ 12 ) we use the previous values (taking in consideration the limits, according to the problem statement )
Z ≤ 10 Z ⇒( 10-9 ) ÷ 1.2 Z = 0.7967
Z ≥ 12 Z ⇒ ( 12 - 9 ) ÷ 1.2 Z = 0.9952
The interval is between these two points
0.9952 - 0.7967 = 0.1985 ⇒ or 19.85 %
The attached help in the understanding of the solution
Answer:
(a): 0.7967
(b): 0.0062
(c): 0.1771
Step-by-step explanation:
Solution :
Given that ,
mean = \mu = 9
standard deviation = \sigma = 1.2
(a)
P(x < 10) = P[(x - \mu ) / \sigma < (10 - 9) / 1.2]
= P(z < 0.83)
= 0.7967
Probability = 0.7967
(b)
P(x > 12) = 1 - P(x < 12)
= 1 - P[(x - \mu ) / \sigma < (12 - 9) / 1.2]
= 1 - P(z < 2.5)
= 1 - 0.9938
= 0.0062
Probability = 0.0062
(c)
P(10 < x < 12) = P[(10 - 9)/ 1.2) < (x - \mu ) /\sigma < (12 - 9) / 1.2) ]
= P(0.83 < z < 2.5)
= P(z < 2.5) - P(z < 0.83)
= 0.9938 - 0.7967
= 0.1971
Probability = 0.1771
