Answer:
the probability that five randomly selected students will have a mean score that is greater than the mean achieved by the students = 0.0096
Step-by-step explanation:
From the five randomly selected students ; 160, 175, 163, 149, 153
mean average of the students = 160+175+163+149+153/5
= mean = x-bar = 800/5
mean x-bar = 160
from probability distribution, P(x-bar > 160) = P[ x-bar - miu / SD > 160 -150.8 /3.94]
P( Z>2.34) = from normal Z-distribution table
= 0.0096419
= 0.0096
hence the probability that five randomly selected students will have a mean score that is greater than the mean achieved by the students = 0.0096
where SD = standard deviation = 3.94 and Miu = 150.8
The answer is isosceles and i think it's acute
3f + 2g = 30 . . .(1)
f + 2g = 26 . . . .(2)
(1) - (2) => 2f = 4 => f = 4/2 = 2
From (2), 2 + 2g = 26 => 2g = 26 - 2 = 24 => g = 24/2 = 12
Therefore, f = 2 and g = 12
Answer:
a. P= 0.6364
b. P = 0.3636
c. Q = $21.25
d. P = 0.5
Step-by-step explanation:
given data
value of a stock varies = $13 to $24
solution
P (stock value is more than $17)
P = 
P = 
P = 0.6364
and
P (value of the stock is between $17 and $21)
P = 
P = 
P = 0.3636
and
Let the upper quartile be Q
(24 - Q) = 2.75
so
Q = $21.25
and
P(X > 20 | X > 16)
P = 
P = 
P = 0.5
