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Chapter 6: Problem 42

Let \(z\) denote a random variable that has a standard normal distribution. Determine each of the following probabilities: a. \(P(z<2.36)\) b. \(P(z \leq 2.36)\) c. \(P(z<-1.23)\) d. \(P(1.142)\) g. \(P(z \geq-3.38)\) h. \(P(z<4.98)\)

Short Answer

Expert verified
The probabilities for \(P(z<2.36)\) and \(P(z \leq 2.36)\) are 0.9910. The probability for \(P(z<-1.23)\) is 0.1093. The probability for \(1.142)\) is 0.0228. The probability for \(P(z \geq -3.38)\) is 0.0003. The probability for \(P(z<4.98)\) is approximately 1.

Step by step solution

01

Determining the probabilities

Firstly, recall that the z-score of a standard normal distribution N(0,1) is already given. To determine the probabilities, use the z-table: these are normally given, showing the probabilities of each z-score. The z-scores have already been found for each of these probabilities.
02

Calculate probability for \(P(z

Locate 2.36 inside the table. Follow that row until you hit the column labeled '.00'. The intersection is the probability answer, 0.9910.
03

Calculate probability for \(P(z \leq 2.36)\)

This probability is the same as \(P(z<2.36)\) because the distribution is continuous, thus including the endpoint doesn't change the probability. Hence, the probability is also 0.9910.
04

Calculate probability for \(P(z

Because the standard normal distribution is symmetric around 0, \(P(z<-1.23)\) will correspond to \(1-P(z<1.23)\). So, find the probability of 1.23 in the z-table, which is 0.8907. Therefore, the probability of \(P(z<-1.23)\) is \(1-0.8907 = 0.1093\).
05

Calculate probability for \(P(1.14

To find this probability, find the probabilities of 3.35 and 1.14 in the z-table, which are 0.9996 and 0.8729 respectively. Subtract these two probabilities to find the probability between these two z-scores: \(0.9996 - 0.8729 = 0.1267\).
06

Calculate probability for \(P(-0.77 \leq z \leq -0.55)\)

Find the probabilities of \(-0.55\) and \(-0.77\) in the z-table, which are 0.7054 and 0.2206 respectively. Subtract these two probabilities to find the required probability: \(0.7054 - 0.2206 = 0.4848\).
07

Calculate probability for \(P(z>2)\)

This corresponds to \(1-P(z<2)\). So, find the probability of 2 in the z-table, which is 0.9772. Therefore, the probability of \(P(z>2)\) is \(1-0.9772 = 0.0228\).
08

Calculate probability for \(P(z \geq -3.38)\)

This corresponds to \(1-P(z<-3.38)\) because of the symmetry of the standard normal distribution. Find the probability of -(-3.38) in the z-table, which is 0.9997. Therefore, the probability of \(P(z \geq -3.38)\) is \(1-0.9997 = 0.0003\).
09

Calculate probability for \(P(z

For values above 3 in the z-table, the probabilities are essentially 1, so we can assume that \(P(z < 4.98)\) is approximately 1.

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Most popular questions from this chapter

Suppose that your statistics professor tells you that the scores on a midterm exam were approximately normally distributed with a mean of 78 and a standard deviation of 7 . The top \(15 \%\) of all scores have been designated A's. Your score is 89. Did you receive an A? Explain.

Suppose that \(25 \%\) of the fire alarms in a large city are false alarms. Let \(x\) denote the number of false alarms in a random sample of 100 alarms. Approximate the following probabilities: a. \(P(20 \leq x \leq 30)\) b. \(P(20 < x < 30)\) c. \(P(x \geq 35)\) d. The probability that \(x\) is farther than 2 standard deviations from its mean value.

Consider babies born in the "normal" range of \(37-43\) weeks gestational age. The paper referenced in Example 6.21 ("Fetal Growth Parameters and Birth Weight: Their Relationship to Neonatal Body Composition," Ultrasound in Obstetrics and Gynecology [2009]: \(441-446\) ) suggests that a normal distribution with mean \(\mu=3,500\) grams and standard deviation \(\sigma=600\) grams is a reasonable model for the probability distribution of \(x=\) birth weight of a randomly selected full-term baby. a. What is the probability that the birth weight of a randomly selected full- term baby exceeds \(4,000 \mathrm{~g} ?\) is between 3,000 and \(4,000 \mathrm{~g}\) ? b. What is the probability that the birth weight of a randomly selected full- term baby is either less than \(2,000 \mathrm{~g}\) or greater than \(5,000 \mathrm{~g}\) ? c. What is the probability that the birth weight of a randomly selected full- term baby exceeds 7 pounds? (Hint: \(1 \mathrm{lb}=453.59 \mathrm{~g} .)\) d. How would you characterize the most extreme \(0.1 \%\) of all full-term baby birth weights?

The light bulbs used to provide exterior lighting for a large office building have an average lifetime of 700 hours. If lifetime is approximately normally distributed with a standard deviation of 50 hours, how often should all the bulbs be replaced so that no more than \(20 \%\) of the bulbs will have already burned out?

Industrial quality control programs often include inspection of incoming materials from suppliers. If parts are purchased in large lots, a typical plan might be to select 20 parts at random from a lot and inspect them. Suppose that a lot is judged acceptable if one or fewer of these 20 parts are defective. If more than one part is defective, the lot is rejected and returned to the supplier. Find the probability of accepting lots that have each of the following (Hint: Identify success with a defective part): a. \(5 \%\) defective parts b. \(10 \%\) defective parts c. \(20 \%\) defective parts
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