Download Solved Problems in Probability and Statistics: Binomial and Normal Distributions and more Exercises Statistics in PDF only on Docsity!
A barber expects to get between zero and six customers per hour in his barber shop. The probability of these is given as follows: Find the number of expected customers that the barber will get per hour. Also, find the variance and standard deviation of this data. The expected value is given by So, the barber can expect 2.46 customers per hour. The variance is given by The standard deviation is given by: A baseball player has a batting average of .211 (in other words, he gets a hit 21.1 % of the time that he goes up to bat). If he goes up to bat 13 times, what is the probability that he will get exactly 4 hits? Here we have n=13, x=4, and p=.211.
A large shipment of light bulbs has just arrived at a store. It has been revealed that 17 % of the light bulbs are defective (the other light bulbs are good). Suppose that you choose 6 light bulbs at random. What is the probability that 2 or less of the bulbs are defective. For two or less, we must calculate the probability of getting two defective, one defective, and zero defective, then add the probabilities. For two defective, we have n=6, x=2, and p=.17. For one defective, we have n=6, x=1, and p=.17. For zero defective, we have n=6, x=0, and p=.17. So, the probability of two, one, or zero defective: .2057+.40178+.3269=.93438.
1. Find P( Z ≤ 1.93). From the table, we get P( Z ≤ 1.93) = .97320.
2. Find P( Z ≤ - .05). From the table, we get P( Z ≤ - .05) = .48006.
(Remember: You may find the standard normal table by clicking Help Files.)
1. Find P( Z ≥ 2.06). P( Z ≤ 2.06). is given directly in the standard normal table and
is found to be .98030. So,
P( Z ≥2.06) = 1 - P ( Z ≤ 2.06) = 1 -.98030 =.
2. Find P( Z ≥ -.78). P( Z≤ -.78). is given directly in the standard normal table and is
found to be .21770. So,
P( Z ≥ - .78) = 1 - P( Z ≤ - .78) = 1 - .21770 = .7823.
3. Find P(-2.20 ≤ Z ≤ -.63). Again, we have:
P(-2.20 ≤ Z ≤ -.63) = P( Z ≤ -.63) - P( Z ≤ -2.20).
Reading the values in the table:
P(-2.20 ≤ Z ≤ -.63) = .26435 - .01390 =.
The time that people stand in the waiting line at a particular fast food restaurant is normally distributed with a mean time of 130 seconds and a standard deviation of 25 seconds. If you go to that fast food restaurant, what is the probability that you will stand in the waiting line for: a) More than 150 seconds? b) Less than 120 seconds? c) Between 110 seconds and 150 seconds? We have μ=130 and σ=25. a) We must find the z-score for x=150:
So, we want P( Z ≥ .80) Since this is greater than, we must use:
b) We must find the z-score for x=120:
So, we want P( Z ≤ -.4). From the table, we find. P( Z ≤ -.4) = .34458.
c) We must find the z-score for x=150: and the z-score for x=110: So, we want
In a large shipment of clocks, it has
been discovered that 21 % of the
clocks are defective. Suppose
that you choose 7 clocks at
random. What is the probability
that 2 or less of the clocks are
defective
In a large shipment of clocks, it has
been discovered that 21 % of the
clocks are defective. Suppose
that you choose 7 clocks at
random. What is the probability
that 2 or less of the clocks are
defective.
In a large shipment of clocks, it has
been discovered that 21 % of the
clocks are defective. Suppose
An archer is shooting arrows at a target. She hits the target 63% of the time. If she takes 19 shots at the target, what is the probability that she will hit the target exactly 11 times? The variance is given by
