A Restaurant hostess is paid $50 plus 10% of the waitstaff's tips for each night she works. If y represents her pay each night a
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Answer:
(a) Probability that a single randomly selected element X of the population is between 57,000 and 58,000 = 0.46411
(b) Probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = 0.99621
Step-by-step explanation:
We are given that a normally distributed population has mean 57,800 and standard deviation 75, i.e.; = 57,800 and
= 750.
Let X = randomly selected element of the population
The z probability is given by;
Z = ~ N(0,1)
(a) So, P(57,000 <= X <= 58,000) = P(X <= 58,000) - P(X < 57,000)
P(X <= 58,000) = P( <=
) = P(Z <= 0.27) = 0.60642
P(X < 57000) = P( <
) = P(Z < -1.07) = 1 - P(Z <= 1.07)
= 1 - 0.85769 = 0.14231
Therefore, P(31 < X < 40) = 0.60642 - 0.14231 = 0.46411 .
(b) Now, we are given sample of size, n = 100
So, Mean of X, X bar = 57,800 same as before
But standard deviation of X, s = =
= 75
The z probability is given by;
Z = ~ N(0,1)
Now, probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = P(57,000 < X bar < 58,000)
P(57,000 <= X bar <= 58,000) = P(X bar <= 58,000) - P(X bar < 57,000)
P(X bar <= 58,000) = P( <=
) = P(Z <= 2.67) = 0.99621
P(X < 57000) = P( <
) = P(Z < -10.67) = P(Z > 10.67)
This probability is that much small that it is very close to 0
Therefore, P(57,000 < X bar < 58,000) = 0.99621 - 0 = 0.99621 .