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Question
- on average the weather in death valley, california averages 119 degrees fahrenheit during the month of july. the standard deviation is 3.8 degrees.
a. what percentage days were 118 degrees or less?
b. what percentage of days were 125 degrees or more?
c. what percent of days was the weather between 118 and 125 degrees in july?
- beaches resorts are family vacation destinations in the caribbean. on average if you go to turks and caicos beaches the average family price is $4,320 with a standard deviation of $350. if you go to antigua beaches, the average family price is $5,600 with a standard deviation of $600. the garcia family went to turks and calcos and spent $4,800. the rodriguez family went to antigua and spent $5,500. explain who received a better deal.
Problem 11a
Step1: Calculate z - score
The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 118\), \(\mu=119\), and \(\sigma = 3.8\).
Step2: Find the percentage from the standard normal table
Looking up \(z=-0.26\) in the standard - normal table (the cumulative distribution function of the standard normal distribution \(\varPhi(z)\)), we find that \(\varPhi(-0.26)=0.3974\)
Step1: Calculate z - score
Using the formula \(z=\frac{x-\mu}{\sigma}\), with \(x = 125\), \(\mu = 119\), and \(\sigma=3.8\)
Step2: Find the percentage
We know that \(P(X\geq x)=1 - P(X\lt x)\). From the standard - normal table, \(P(Z\lt1.58) = 0.9429\). So \(P(Z\geq1.58)=1 - 0.9429=0.0571\)
Step1: Calculate z - scores
For \(x = 118\): \(z_1=\frac{118 - 119}{3.8}\approx - 0.26\)
For \(x = 125\): \(z_2=\frac{125 - 119}{3.8}\approx1.58\)
Step2: Find the probabilities
From the standard - normal table, \(P(Z\lt - 0.26)=0.3974\) and \(P(Z\lt1.58)=0.9429\)
Then \(P(-0.26\lt Z\lt1.58)=P(Z\lt1.58)-P(Z\lt - 0.26)\)
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\(39.74\%\)