Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Find the following probabilities for the standard normal random variable: a. \(P(.3

Short Answer

Expert verified
Answer: The probabilities for the given standard normal random variable are: a. \(P(.3<z<1.56) = 0.3227\) b. \(P(-.2<z<.2) = 0.1586\)

Step by step solution

01

Use the Z-table for the first interval (\(P(.3

To find the probabilities for the first interval, refer to the Z-table to look up the probabilities for \(z = .3\) and \(z = 1.56\). Based on the Z-table, we get: - \(P(z \le .3) = 0.6179\) - \(P(z \le 1.56) = 0.9406\)
02

Calculate the probability for the first interval

Now, we will subtract the probability for the lower limit from the probability for the upper limit to find the probability between these points: \(P(.3 < z < 1.56) = P(z \le 1.56) - P(z \le .3)\) \(P(.3 < z < 1.56) = 0.9406 - 0.6179 = 0.3227\) The probability for the first interval is 0.3227.
03

Use the Z-table for the second interval (\(P(-.2

Repeat the process for the second interval. Refer to the Z-table to look up the probabilities for \(z = -.2\) and \(z = .2\). Based on the Z-table, we get: - \(P(z \le -.2) = 0.4207\) - \(P(z \le .2) = 0.5793\)
04

Calculate the probability for the second interval

Again, subtract the probability for the lower limit from the probability for the upper limit to find the probability between these points: \(P(-.2 < z < .2) = P(z \le .2) - P(z \le -.2)\) \(P(-.2 < z < .2) = 0.5793 - 0.4207 = 0.1586\) The probability for the second interval is 0.1586. In summary, the probabilities for the given standard normal random variable are: a. \(P(.3<z<1.56) = 0.3227\) b. \(P(-.2<z<.2) = 0.1586\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider a binomial random variable with \(n=45\) and \(p=.05 .\) Fill in the blanks below to find some probabilities using the normal approximation. a. Can we use the normal approximation? Calculate \(n p=\) ______ and \(n q=\) ______ b. Are \(n p\) and \(n q\) both greater than \(5 ?\) Yes_____ No______ c. If the answer to part \(b\) is yes, calculate \(\mu=n p=\) _______ and \(\sigma=\sqrt{n p q}=\) _____ d. To find the probability of 10 or fewer successes, what values of \(x\) should be included? \(x=\) _______ e. To include the entire block of probability for the first value of \(x=\) ______, start at ______. f. Calculate \(z=\frac{x \pm .5-n p}{\sqrt{n p q}}=\) _______. g. Calculate \(P(x \leq 10) \approx P(z<\)_____ ) \(=\) ______.

There is a difference in sports preferences between men and women, according to a recent survey. Among the 10 most popular sports, men include competition- type sports-pool and billiards, basketball, and softball-whereas women include aerobics, running, hiking, and calisthenics. However, the top recreational activity for men was still the relaxing sport of fishing, with \(41 \%\) of those surveyed indicating that they had fished during the year. Suppose 180 randomly selected men are asked whether they had fished in the past year. a. What is the probability that fewer than 50 had fished? b. What is the probability that between 50 and 75 had fished? c. If the 180 men selected for the interview were selected by the marketing department of a sporting goods company based on information obtained from their mailing lists, what would you conclude about the reliability of their survey results?

a. Find a \(z_{0}\) such that \(P\left(z>z_{0}\right)=.9750 .\) b. Find a \(z_{0}\) such that \(P\left(z>z_{0}\right)=.3594\).

a. Find a \(z_{0}\) such that \(P\left(-z_{0}

Students very often ask their professors whether they will be "curving the grades." The traditional interpretation of "curving grades" required that the grades have a normal distribution, and that the grades will be assigned in these proportions: $$ \begin{array}{l|lllll} \text { Letter Grade } & \mathrm{A} & \mathrm{B} & \mathrm{C} & \mathrm{D} & \mathrm{F} \\ \hline \text { Proportion of Students } & 10 \% & 20 \% & 40 \% & 20 \% & 10 \% \end{array} $$ a. If the average "C" grade is centered at the average grade for all students, and if we assume that the grades are normally distributed, how many standard deviations on either side of the mean will constitute the "C" grades? b. How many deviations on either side of the mean will be the cutoff points for the "B" and "D" grades?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free