The average demand for rental skis on winter Saturdays at a particular area is 150 pairs, which has been quite stable over time.
1 answer:
Answer:
a) 15.87% probability that demand will exceed this supply on any winter Saturday
b) The shop needs to have 197 pairs of skis in stock.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
(a) The rental shop stocks 170 pairs of skis. What is the probability that demand will exceed this supply on any winter Saturday?
This probability is 1 subtracted by the pvalue of Z when X = 150. So
has a pvalue of 0.8413
1 - 0.8413 = 0.1587
15.87% probability that demand will exceed this supply on any winter Saturday
(b) How many pairs of skis in stock does the shop have to have to make the probability in question (a) less than .01?
This is X when Z has a pvalue of 0.99.
So X when Z = 2.33
Rounding up, since the number of pairs is a discrete number
The shop needs to have 197 pairs of skis in stock.
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Step-by-step explanation:
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Answer:
95.64% probability that under 30% of the dogs are emotional support trained
Step-by-step explanation:
For each dog, there are only two possible outcomes. Either they have completed emotional support training, or they have not. So we use the binomial probability distribution to solve this problem.
However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that ,
.
In this problem, we have that:
What is the probability that under 30% of the dogs are emotional support trained?
30% of 100 is 0.3*100 = 30
So this is the pvalue of Z when X = 30.
has a pvalue of 0.9564.
So there is a 95.64% probability that under 30% of the dogs are emotional support trained