Answer: did anyone get the answer to this ?
Step-by-step explanation:
Answer:
1. x=40
2. x=17
3. x=5
4. x=83
Step-by-step explanation:
Using the normal distribution, it is found that 0.26% of the items will either weigh less than 87 grams or more than 93 grams.
In a <em>normal distribution</em> with mean 
and standard deviation 
, the z-score of a measure X is given by:
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem:
- The mean is of 90 grams, hence
.
- The standard deviation is of 1 gram, hence
.
We want to find the probability of an item <u>differing more than 3 grams from the mean</u>, hence:
The probability is P(|Z| > 3), which is 2 multiplied by the p-value of Z = -3.
- Looking at the z-table, Z = -3 has a p-value of 0.0013.
2 x 0.0013 = 0.0026
0.0026 x 100% = 0.26%
0.26% of the items will either weigh less than 87 grams or more than 93 grams.
For more on the normal distribution, you can check brainly.com/question/24663213
Answer:
Since the calculated value of z= -1.496 does not fall in the critical region z < -1.645 we conclude that the new program is effective. We fail to reject the null hypothesis .
Step-by-step explanation:
The sample proportion is p2= 7/27= 0.259
and q2= 0.74
The sample size = n= 27
The population proportion = p1= 0.4
q1= 0.6
We formulate the null and alternate hypotheses that the new program is effective
H0: p2> p1 vs Ha: p2 ≤ p1
The test statistic is
z= p2- p1/√ p1q1/n
z= 0.259-0.4/ √0.4*0.6/27
z= -0.141/0.09428
z= -1.496
The significance level ∝ is 0.05
The critical region for one tailed test is z ≤ ± 1.645
Since the calculated value of z= -1.496 does not fall in the critical region z < -1.645 we conclude that the new program is effective. We fail to reject the null hypothesis .
Answer: just follow these simple steps!
Step-by-step explanation:
1. Estimate the quotient.
2. Perform the division. Remember to place a zero in the quotient when the divisor is larger than the dividend. ...
3. Compare your estimate with your quotient to verify that the answer makes sense.
