Question

A boat capsized and sank in a lake. Based on an assumption of a mean weight of

130130

​lb, the boat was rated to carry

7070

passengers​ (so the load limit was

9 comma 1009,100

​lb). After the boat​ sank, the assumed mean weight for similar boats was changed from

130130

lb to

172172

lb. Complete parts a and b below.

a. Assume that a similar boat is loaded with

7070

​passengers, and assume that the weights of people are normally distributed with a mean of

175.6175.6

lb and a standard deviation of

40.840.8

lb. Find the probability that the boat is overloaded because the

7070

passengers have a mean weight greater than

130130

lb.The probability is

nothing.

​(Round to four decimal places as​ needed.)

b. The boat was later rated to carry only

1717

​passengers, and the load limit was changed to

2 comma 9242,924

lb. Find the probability that the boat is overloaded because the mean weight of the passengers is greater than

172172

​(so that their total weight is greater than the maximum capacity of

2 comma 9242,924

​lb).The probability is

nothing.

​(Round to four decimal places as​ needed.)

Do the new ratings appear to be safe when the boat is loaded with

1717

​passengers? Choose the correct answer below.

A.Because there is a high probability of​ overloading, the new ratings do not appear to be safe when the boat is loaded with

1717

passengers.

B.

Because the probability of overloading is lower with the new ratings than with the old​ ratings, the new ratings appear to be safe.

C.Because there is a high probability of​ overloading, the new ratings appear to be safe when the boat is loaded with

1717

passengers.

D.Because

175.6175.6

is greater than

172172​,

the new ratings do not appear to be safe when the boat is loaded with

1717

passengers.

___________________________________________________________________________________________________________________

Assume that females have pulse rates that are normally distributed with a mean of

mu equals 75.0μ=75.0

beats per minute and a standard deviation of

sigma equals 12.5σ=12.5

beats per minute. Complete parts​ (a) through​ (c) below.

a. If 1 adult female is randomly​ selected, find the probability that her pulse rate is between

6868

beats per minute and

8282

beats per minute.The probability is

0.42460.4246.

​(Round to four decimal places as​ needed.)

b. If

1616

adult females are randomly​ selected, find the probability that they have pulse rates with a mean between

6868

beats per minute and

8282

beats per minute.The probability is

00.

​(Round to four decimal places as​ needed.)

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

A.

Since the distribution is of​ individuals, not sample​ means, the distribution is a normal distribution for any sample size.

B.

Since the distribution is of sample​ means, not​ individuals, the distribution is a normal distribution for any sample size.

C.

Since the mean pulse rate exceeds​ 30, the distribution of sample means is a normal distribution for any sample size.

D.

Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

Answer 1

a.

Standard error of mean = 40.8 / = 4.876533

P(X > 130) = P[Z > (130 - 175.6) / 4.876533] = P[Z > -9.35] = 1.0000

b.

Standard error of mean = 40.8 / = 9.895454

P(X > 172) = P[Z > (172 - 175.6) / 9.895454] = P[Z > -0.36] = 0.6406

A.Because there is a high probability of​ overloading, the new ratings do not appear to be safe when the boat is loaded with 17 passengers.

___________________________________________________________________________________________________________________

μ=75.0 beats per minute

σ=12.5 beats per minute.

a.

P(68 < X < 82) = P(X < 82) - P(X < 68) = P[Z < (82 - 75 )/ 12.5] - P[Z < (68 - 75) / 12.5]

= P[Z < 0.56] - P[Z < -0.56]

= 0.7123 - 0.2877

= 0.4246

b.

Standard error of mean = 12.5 / = 3.125

P(68 < < 82) = P( < 82) - P( < 68) = P[Z < (82 - 75 )/ 3.125] - P[Z < (68 - 75) / 3.125]

= P[Z < 2.24] - P[Z < -2.24]

= 0.9875 - 0.0125

= 0.9749

c.

D.

Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.