Question

In: Statistics and Probability

1)Use the normal distribution of SAT critical reading scores for which the mean is 513 and...

1)Use the normal distribution of SAT critical reading scores for which the mean is 513 and the standard deviation is 110. Assume the variable x is normally distributed.

What percent of the SAT verbal scores are less than 550 ?

If 1000 SAT verbal scores are randomly selected, about how many would you expect to be greater than 525?

2, You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Which interval is wider? If convenient, use technology to construct the confidence intervals.

A random sample of 34 gas grills has a mean price of $646.60.

Assume the population standard deviation is $59.30. The 90% confidence interval is (__

The 90% confidence interval is (___,___)

The 95% confidence interval is (__,___)

Which interval is wider? Choose the correct answer below.

The 95% confidence interval

The 90% confidence interval

3, In a recent year, the scores for the reading portion of a test were normally distributed, with a mean of 22.2 and a standard deviation of 6.1. Complete parts (a) through (d) below.

(a) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is less than 18.

The probability of a student scoring less than 18 is (___)

(Round to four decimal places as needed.)

(b) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is between 15.7 and 28.7.

The probability of a student scoring between 15.7 and 28.7 is( )

(c) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is more than 34.7.

Expert Solution

1) a) P(X < 550)

= P((X - )/ < (550 - )/)

= P(Z < (550 - 513)/110)

= P(Z < 0.34)

= 0.6331

b) P(X > 525)

= P((X - )/ > (525 - )/)

= P(Z > (525 - 513)/110)

= P(Z > 0.11)

= 1 - P(Z < 0.11)

= 1 - 0.5438

= 0.4562

Expected value = 1000 * 0.4562 = 456.2 = 456

2) At 90% confidence interval the critical value is z_0.05 = 1.645

The 90% confidence interval is

+/- z_0.05 *

= 646.60 +/- 1.645 * 59.30/sqrt(34)

= 646.60 +/- 16.729

= 629.871, 663.329

At 95% confidence interval the critical value is z_0.025 = 1.96

The 90% confidence interval is

+/- z_0.025 *

= 646.60 +/- 1.96 * 59.30/sqrt(34)

= 646.60 +/- 19.93

= 626.67, 666.53

The 95% confidence interval is wider.

3)a) P(X < 18)

= P((X - )/ < (18 - )/)

= P(Z < (18 - 22.2)/6.1)

= P(Z < -0.69)

= 0.2451

b) P(15.7 < X < 28.7)

= P((15.7 - )/ < (X - )/ < (28.7 - )/)

= P((15.7 - 22.2)/6.1 < Z < (28.7 - 22.2)/6.1)

= P(-1.07 < Z < 1.07)

= P(Z < 1.07) - P(Z < -1.07)

= 0.8577 - 0.1423

= 0.7154

c) P(X > 34.7)

= P((X - )/ > (34.7 - )/)

= P(Z > (34.7 - 22.2)/6.1)

= P(Z > 2.05)

= 1 - P(Z < 2.05)

= 1 - 0.9798

= 0.0202

orchestra answered 2 years ago

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