The talk-time battery life of a group of cell phone is normally distributed with a mean of 5 hours and a standard deviation of 1
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Answer:
P(2.50 < Xbar < 2.66) = 0.046
Step-by-step explanation:
We are given that Population Mean, = 2.58 and Standard deviation,
= 0.75
Also, a random sample (n) of 110 households is taken.
Let Xbar = sample mean household size
The z score probability distribution for sample mean is give by;
Z = ~ N(0,1)
So, probability that the sample mean household size is between 2.50 and 2.66 people = P(2.50 < Xbar < 2.66)
P(2.50 < Xbar < 2.66) = P(Xbar < 2.66) - P(Xbar 2.50)
P(Xbar < 2.66) = P( <
) = P(Z < -1.68) = 1 - P(Z 1.68)
= 1 - 0.95352 = 0.04648
P(Xbar 2.50) = P(
) = P(Z
-3.92) = 1 - P(Z < 3.92)
= 1 - 0.99996 = 0.00004
Therefore, P(2.50 < Xbar < 2.66) = 0.04648 - 0.00004 = 0.046