Answer:
(a) P(
76) = 0.2327
(b) P(73 < < 75) = 0.5035
(c) P( < 74.8) = 0.77035
Step-by-step explanation:
We are given that the mean of a population is 74 and the standard deviation is 16.
Assuming the data follows normal distribution.
<u><em>Let </em></u><u><em> = sample mean </em></u>
The z-score probability distribution for sample mean is given by;
Z = ~ N(0,1)
where, = population mean = 74
= standard deviation = 16
n = sample size
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
(a) Probability that a random sample of size 34 yielding a sample mean of 76 or more is given by = P(
76)
P(
76) = P(
) = P(Z
0.73) = 1 - P(Z < 0.73)
= 1 - 0.7673 = <u>0.2327</u>
<em>The above probability is calculated by looking at the value of x = 0.73 in the z table which has an area of 0.7673.</em>
<em />
(b) Probability that a random sample of size 120 yielding a sample mean of between 73 and 75 is given by = P(73 < < 75) = P(
< 75) - P(
73)
P( < 75) = P(
<
) = P(Z < 0.68) = 0.75175
P(
73) = P(
) = P(Z
-0.68) =1 - P(Z < 0.68)
= 1 - 0.75175 = 0.24825
Therefore, P(73 < < 75) = 0.75175 - 0.24825 = <u>0.5035</u>
<em>The above probability is calculated by looking at the value of x = 0.68 in the z table which has an area of 0.75175.</em>
<em />
(c) Probability that a random sample of size 218 yielding a sample mean of less than 74.8 is given by = P( < 74.8)
P( < 74.8) = P(
<
) = P(Z < 0.74) = <u>0.77035</u>
<em>The above probability is calculated by looking at the value of x = 0.74 in the z table which has an area of 0.77035.</em>