Question

TIUNTIVOIR. DISLUULUMUL UUMPIC IVICUS VVCCR IyU." Score: 0 of 1 pt 8 of 11 (10 complete) HW Score: 71.21%, 7.83 of 11 pts X 8.1.21 Question Help The reading speed of second grade students in a large city is approximately normal, with a mean of 92 words per minute (wpm) and a standard deviation of 10 wpm. Complete parts (a) through (n. Click here to view the standard normal distribution table (page 1) Click here to view the standard normal distribution table page 2 (a) What is the probability a randomly selected student in the city will read more than 96 words per minuto? The probability is 0.3446 (Round to four decimal places as needed.) Interpret this probability. Select the correct choice below and fill in the answer box within your choice A 100 different students were chosen from this population, we would expect 100 different students were chosen from this population, we would expect to read exactly 6 words per minute to read less than 96 words per minute If 100 different students were chosen from this population, we would expect 34 to read more than 96 words per minute (b) What is the probability that a random sample of 12 second grade students from the city results in a mean reading rate of giore than 96 words per minute? The probability is 0.0829 (Round to four decimal places as needed) Interpret this probability. Select the correct choice below and fill in the answer box within your choice A 100 different samples of n=12 students were chosen from this population, we would expect sample(s) to have a sample mean reading rate of less than 96 words per minute. 3. If 100 different samples of n 12 students were chosen from this population, we would expect sample(s) to have a sample mean reading rate of exactly 96 words per minute *C 100 different samples of n = 12 students were chosen from this population, we would expect 8 sample(s) to have a sample mean reading rate of more than 96 words per minute (c) What is the probability that a random sample of 24 second grade students from the city results in a mean reading rate of more than words per minute? The probability is 0.0250 (Round to four decimal places as needed) Interprets probability. Select the correct choice below and in the newer box within your choice Question is complete. Tap on the red indicators to see incorrect answers. All parts showing Similar Question

please help! NO STANDARD NORMAL DISTRIBUTION TABLE. please use normalcdf or formula. Thank you

Answer 1

Here we have

$\mu=92,\sigma=10$

(a)

The z-score for x = 96 is

$z=\frac{96-92}{10}=0.40$

So the required probability is

$P(x>96)=1-P(x\leq 96)=P(z>0.4)=1-P(z\leq 0.4)=0.3445$

Expected number of students = 100 * 0.3445= 34

Following is the screen shot of calculator:

1-normalcdf<-999 -0.4,0,1)) 3445783629

(b)

Sample size n=12

The z-score for $\bar{x}=96$ is

$z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\frac{96-92}{10/\sqrt{12}}=1.39$

So required probability is

$P(\bar{x}>96)=P(z>1.39)=1-P(z\leq 1.39)=0.0823$

Expected number of students = 100 * 0.0823= 8

Following is the screen shot:

1.39,021644925 normalçaf<-999

(c)

Sample size n=24

The z-score for $\bar{x}=96$ is

$z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\frac{96-92}{10/\sqrt{24}}=1.96$

So required probability is

$P(\bar{x}>96)=P(z>1.96)=1-P(z\leq 1.96)=0.0250$

Expected number of students = 100 * 0.0250=3

Following is the screen shot:

1-normalcdf<-999 -1.96,0,1) :6249978252