Quantitative Methods Midterm 2

70-60 /8 =1.25 = 0.8944 38-60 /8= -2.75 = 0.0030 .8944-.0030 = .8914 C. .8914

Assume that the random variable X is normally distributed, with mean =60 and standard deviation =8 Compute the probability P(38<x<70)

75-60 /15 =0.8 =.7881 1-.7881 =.2119

Assume that the random variable X is normally distributed, with mean MU=80 and standard deviation = 15 Compute the probability P(X>92)

C, No the graph is skewed right

Determine the whether the graph could represent a variable with normal distribution. Explain your reasoning. High points in the beginning of the graph...

A) 0.9032 1-.0968= .0932

Find the area indicated region under the standard normal curve -1.3 and to the right

-0.45 = .3264 2.11= .9826 .9826-.3264= 0.6562 D) 0.6562

Find the area of the indicated region under the standard normal curve 2 midpoints: -0.45 and 2.11

C) .9332

Find the area under the standard normal curve to the left of z= 1.5

look inside table for the closest to 0.75, at .7486 is z= 0.67 C. 0.67

For the standard normal curve, find the z-score that corresponds to the 3rd quartile

110-100 /15 =.67 C) .67

IQ test scores are normally distributed with a mean of 100 and a standard deviation of 15. an individuals iq score is found to be 110. find the z-score corresponding to this value

Student 1: 1930-1475 /308= 1.477 Student 2: 1340-1475 /308= 0.438 Student 3: 2150-1475 /308= 2.19 Student 4: 1450-1475 /308= 0.081

The SAT is an exam used by colleges to evaluate undergraduate applicants. The test scores are normally distributed. in a recent year, the mean test score was 1475 and the standard deviation was 308. The test scores of 4 students selected at random are 1930, 1340, 2150, 1450. Find z score

at -2.5 = 0.0062. at 1.5= .9332 .9332-.0062= .927 A) .9270

Use the standard normal distribution to find P (-2.5 <z <1.5)

20- 15.5 / 3.6 1.25 =.8944 10-15.5 /3.6 =-1.53 -1.53 =.0630 .8944-.0630 = .8314 D. 0.8314

an airline knows from experience that the distribution of the number of suitcases that get lost each week on a certain route is approximately normal with MU=15.5 and std dev= 3.6 what is the probability that during a given week the airline will lose between 10 and 20 suitcases

at -2.33= 0.0099 at 2.33- 0.0099 .0099+.0099= 0.0198 D) 0.0198

use the standard normal distribution to find P(z < -2.33 or z> 2.33)

M=268 Sigma= 15 217-268 /15 -1.4 = .0808x100 =8.08

the lengths of the pregnancies of humans are normally distributed with a mean of 268 days and a standard deviation of 15 days. a baby is premature if its born 3 weeks early. what percent of babies are born prematurely

the two scores are statistically the same 75-65 /8 =1.25 75-70 /4 =1.25

compare the scores: a score of 75 on a test with a mean of 65 and a standard deviation of 8 and a score of 75 on a test with a mean of 70 and a standard deviation of 4

find probability of 0.01 closest one is .0099 and z score is -2.33, 2.33

find the score for which 98% of the distribution area lies between -z and z

1-.7= .3 inside table is 0.3. closest one is 0.2981 and .3015 =-.47

find the z score for which 70% of the distributions area lies to the right

-0.58

find the z score that corresponds to the given area under the standard normal curve 0.2810 to the left

4.2

find the z score that corresponds to the given area under the standard normal curve 0.6628 to the left

-1.71 1-.9564= 0.0436 0.0436 inside table = z score of -1.71

find the z score that corresponds to the given area under the standard normal curve.

.3 on z table = -.052

for the standard normal curve, find the z score that corresponds to the 30th percentile

p(x<78) = 0.0548 -1.6 = 78-MU =86

in a certain normal distribution, find the mean when std dev =5 and 5.48% of the area lies to the left of 78

2.33 = X-100 /15 x=134.95

iq test scores are normally distributed with a mean of 100 and a standard deviation of 15. find the x score that corresponds to the z score of 2.33

-1.645= MU-100 /15 .0505-100 /15 = -6.66 =75.3

iq test scores are normally distributed with a mean of 100 and a standard deviation of 15. find the x-score that corresponds to the z score of -1.645

300-268 / 15= 2.13 1-.9834 = .0166 A) 0.0166

the lengths of pregnancies of humans are normally distributed with a mean of 268 days and a standard deviation of 15 days. find the probability of a pregnancy lasting more than 300 days