John spent $36 to buy 4

The number of chocolate chips in an 18-ounce bag of chocolate chip cookies is approximately normally distributed with mean 1252

stich3 [128]

Answer:

a) P ( 1100 < X < 1400 ) = 0.755

b) P ( X < 1000 ) = 0.755

c) proportion ( X > 1200 ) = 65.66%

d) 5.87% percentile

Step-by-step explanation:

Solution:-

- Denote a random variable X: The number of chocolate chip in an 18-ounce bag of chocolate chip cookies.

- The RV is normally distributed with the parameters mean ( u ) and standard deviation ( s ) given:

u = 1252

s = 129

- The RV ( X ) follows normal distribution:

X ~ Norm ( 1252 , 129^2 )

a) what is the probability that a randomly selected bag contains between 1100 and 1400 chocolate chips?

- Compute the standard normal values for the limits of required probability using the following pmf for standard normal:

P ( x1 < X < x2 ) = P ( [ x1 - u ] / s < Z < [ x2 - u ] / s )

- Taking the limits x1 = 1100 and x2 = 1400. The standard normal values are:

P ( 1100 < X < 1400 ) = P ( [ 1100 - 1252 ] / 129 < Z < [ 1400 - 1252 ] / 129 )

= P ( - 1.1783 < Z < 1.14728 )

- Use the standard normal tables to determine the required probability defined by the standard values:

P ( -1.1783 < Z < 1.14728 ) = 0.755

Hence,

P ( 1100 < X < 1400 ) = 0.755 ... Answer

b) what is the probability that a randomly selected bag contains fewer than 1000 chocolate chips?

- Compute the standard normal values for the limits of required probability using the following pmf for standard normal:

P ( X < x2 ) = P ( Z < [ x2 - u ] / s )

- Taking the limit x2 = 1000. The standard normal values are:

P ( X < 1000 ) = P ( Z < [ 1000 - 1252 ] / 129 )

= P ( Z < -1.9535 )

- Use the standard normal tables to determine the required probability defined by the standard values:

P ( Z < -1.9535 ) = 0.0254

Hence,

P ( X < 1000 ) = 0.755 ... Answer

(c) what proportion of bags contains more than 1200 chocolate chips?

- Compute the standard normal values for the limits of required probability using the following pmf for standard normal:

P ( X > x1 ) = P ( Z > [ x1 - u ] / s )

- Taking the limit x1 = 1200. The standard normal values are:

P ( X > 1200 ) = P ( Z > [ 1200 - 1252 ] / 129 )

= P ( Z > 0.4031 )

- Use the standard normal tables to determine the required probability defined by the standard values:

P ( Z > 0.4031 ) = 0.6566

Hence,

proportion of X > 1200 = P ( X > 1200 )*100 = 65.66% ... Answer

d) what is the percentile rank of a bag that contains 1050 chocolate chips?

- The percentile rank is defined by the proportion of chocolate less than the desired value.

- Compute the standard normal values for the limits of required probability using the following pmf for standard normal:

P ( X < x2 ) = P ( Z < [ x2 - u ] / s )

- Taking the limit x2 = 1050. The standard normal values are:

P ( X < 1050 ) = P ( Z < [ 1050 - 1252 ] / 129 )

= P ( Z < 1.5659 )

- Use the standard normal tables to determine the required probability defined by the standard values:

P ( Z < 1.5659 ) = 0.0587

Hence,

Rank = proportion of X < 1050 = P ( X < 1050 )*100

= 0.0587*100 %

= 5.87 % ... Answer

6 0

3 years ago

if a product is it in late you receive 8/9 of you earn points you receive 72 points on your Late project how many points did you

igor_vitrenko [27]

Hello! So if your project is turned in late, you get 8/9 of the points you would've received. In this case, you only received 72 points. To find out how many points you lost, first, let's find the score you could've gotten if it was turned in on time. Let's set up a proportion like this: 72/x = 8/9. Cross multiply the values to get 648 = 8x. Divide by 8 to isolate the "x" and you get x = 81. You could've originally gotten an 81. However, you only got a 72. To find out the amount of points you lost, subtract 72 from 81. 81 - 72 = 9. You lost 9 points from your original grade.

3 0

3 years ago

The resting heart rates for 80 women aged 46–55 in a simple random sample are normally distributed, with a mean of 71 beats per

salantis [7]

ME = 1.10.

Explanation:
The margin of error, ME, is given by the formula z*(σ/√n).

Since we want a 90% confidence interval, z=1.645.

σ=6, and n=80, giving us ME=1.645*(6/√80). This comes out to 1.10.

7 0

3 years ago

Read 2 more answers

There are 15 male teachers in the school. There are 35 female teachers in the school. Fill in the table

yulyashka [42]

Where is the table???

3 0

3 years ago

a math teacher claims that she has developed a review course that increases the score of students on the math portion of a colle

madam [21]

Answer:

A.

$H_0: \mu\leq514\\\\H_1: \mu>514$

B. Z=2.255. P=0.01207.

C. Although it is not big difference, it is an improvement that has evidence. The scores are expected to be higher in average than without the review course.

D. P(z>0.94)=0.1736

Step-by-step explanation:

A. state the null and alternative hypotheses.

The null hypothesis states that the review course has no effect, so the scores are still the same. The alternative hypothesis states that the review course increase the score.

$H_0: \mu\leq514\\\\H_1: \mu>514$

B. test the hypothesis at the a=.10 level of confidence. is a mean math score of 520 significantly higher than 514? find the test statistic, find P-value. is the sample statistic significantly higher?

The test statistic Z can be calculated as

$Z=\frac{M-\mu}{s/\sqrt{N}} =\frac{520-514}{119/\sqrt{2000}}=\frac{6}{2.661}=2.255$

The P-value of z=2.255 is P=0.01207.

The P-value is smaller than the significance level, so the effect is significant. The null hypothesis is rejected.

Then we can conclude that the score of 520 is significantly higher than 514, in this case, specially because the big sample size.

C. do you think that a mean math score of 520 vs 514 will affect the decision of a school admissions adminstrator? in other words does the increase in the score have any practical significance?

Although it is not big difference, it is an improvement that has evidence. The scores are expected to be higher in average than without the review course.

D. test the hypothesis at the a=.10 level of confidence with n=350 students. assume that the sample mean is still 520 and the sample standard deviation is still 119. is a sample mean of 520 significantly more than 514? find test statistic, find p value, is the sample mean statisically significantly higher? what do you conclude about the impact of large samples on the p-value?

In this case, the z-value is

$Z=\frac{520-514}{s/\sqrt{n}} =\frac{6}{119/\sqrt{350}} =\frac{6}{6.36} =0.94\\\\P(z>0.94)=0.1736>\alpha$

In this case, the P-value is greater than the significance level, so there is no evidence that the review course is increasing the scores.

The sample size gives robustness to the results of the sample. A large sample is more representative of the population than a smaller sample. A little difference, as 520 from 514, but in a big sample leads to a more strong evidence of a change in the mean score.

Large samples give more extreme values of z, so P-values that are smaller and therefore tend to be smaller than the significance level.

8 0

3 years ago