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Chapter 7: Problem 64

Dartmouth Distribution Warehouse makes deliveries of a large number of products to its customers. It is known that \(85 \%\) of all the orders it receives from its customers are delivered on time. Let \(\hat{p}\) be the proportion of orders in a random sample of 100 that are delivered on time. Find the probability that the value of \(\hat{p}\) will be a. between \(.81\) and \(.88\) b. less than \(.87\)

Short Answer

Expert verified
a. The probability that the proportion of orders in a random sample of 100 that are delivered on time is between 0.81 and 0.88 is approximately 0.656.\n b. The probability that the proportion is less than 0.87 is approximately 0.7088.

Step by step solution

01

Calculate mean and standard deviation

The mean (\(碌\)) of a binomial distribution can be calculated as \(碌=np\) and the standard deviation (\(蟽\)) can be found with \(\sqrt{np(1-p)}\) . Here, \(n=100\) and \(p=0.85\), so \(碌=100 * 0.85 = 85\) and \(蟽=\sqrt{100 * 0.85 * 0.15} \approx 3.67\).
02

Calculate z-scores

Z-score is calculated using:\(Z= \frac{x-碌}{蟽}\). Where \(x\) is the sample value, \(碌\) is the mean, and \(蟽\) is the standard deviation. For our first scenario where the sample proportion lies between 0.81 and 0.88, we calculate 2 z-scores: Firstly, \(Z1= \frac{81-85}{3.67} \approx -1.09\) for 0.81 and secondly, \(Z2= \frac{88-85}{3.67} \approx 0.82\) for 0.88.
03

Find probabilities for the first scenario

Normally, we would take these Z-values and look up their corresponding probabilities from a Z-table. From the Z-table, we find that the probability for \(Z1\) is \(0.1379\) and for \(Z2\) is \(0.7939\). As we are looking for probabilities between 0.81 and 0.88, the overall probability is the difference between these two probabilities, i.e., \(P = 0.7939 - 0.1379 = 0.656\).
04

Calculate Z-score for second scenario

For our second scenario where the sample proportion is less than 0.87, we calculate another z-score: \(Z= \frac{87-85}{3.67} \approx 0.55\).
05

Find probabilities for the second scenario

Again from the Z-table, we find the probability associated with this Z-score is \(0.7088\). So, the probability that the sample proportion will be less than 0.87 is \(0.7088\).

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