Math / Data & Statistics

QuestionDetermine whether the normal sampling distribution cain be used. The claim is $p<0.015$ and the sample size is $n=150$ . Use the normal distribution. Do not use the normal distribution.

Studdy Solution

STEP 1

1. The sample proportion is denoted by $\hat{p}$ .
2. The population proportion claim is $p < 0.015$ .
3. The sample size is $n = 150$ .
4. The normal approximation can be used if both $np$ and $n(1-p)$ are greater than or equal to 5.

STEP 2

1. Calculate $np$ and $n(1-p)$ .
2. Evaluate whether the normal distribution can be used based on the calculated values.

STEP 3

Calculate $np$ :
$np = 150 \times 0.015 = 2.25$

STEP 4

Calculate $n(1-p)$ :
$n(1-p) = 150 \times (1 - 0.015) = 150 \times 0.985 = 147.75$

STEP 5

Evaluate whether the normal distribution can be used:
- $np = 2.25$ is less than 5. - $n(1-p) = 147.75$ is greater than 5.
Since $np < 5$ , the normal distribution cannot be used.
The conclusion is:
Do not use the normal distribution.

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QuestionDetermine whether the normal sampling distribution cain be used. The claim is $p<0.015$ and the sample size is $n=150$ . Use the normal distribution. Do not use the normal distribution.

Studdy Solution

STEP 1

1. The sample proportion is denoted by $\hat{p}$ .
2. The population proportion claim is $p < 0.015$ .
3. The sample size is $n = 150$ .
4. The normal approximation can be used if both $np$ and $n(1-p)$ are greater than or equal to 5.

STEP 2

1. Calculate $np$ and $n(1-p)$ .
2. Evaluate whether the normal distribution can be used based on the calculated values.

STEP 3

Calculate $np$ :
$np = 150 \times 0.015 = 2.25$

STEP 4

Calculate $n(1-p)$ :
$n(1-p) = 150 \times (1 - 0.015) = 150 \times 0.985 = 147.75$

STEP 5

Evaluate whether the normal distribution can be used:
- $np = 2.25$ is less than 5. - $n(1-p) = 147.75$ is greater than 5.
Since $np < 5$ , the normal distribution cannot be used.
The conclusion is:
Do not use the normal distribution.

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QuestionDetermine whether the normal sampling distribution cain be used. The claim is p<0.015p<0.015p<0.015 and the sample size is n=150n=150n=150. Use the normal distribution. Do not use the normal distribution.

Studdy Solution

STEP 1

1. The sample proportion is denoted by p^ \hat{p} p^​.2. The population proportion claim is p<0.015 p < 0.015 p<0.015.3. The sample size is n=150 n = 150 n=150.4. The normal approximation can be used if both np np np and n(1−p) n(1-p) n(1−p) are greater than or equal to 5.

STEP 2

1. Calculate np np np and n(1−p) n(1-p) n(1−p).2. Evaluate whether the normal distribution can be used based on the calculated values.

STEP 3

Calculate np np np:np=150×0.015=2.25 np = 150 \times 0.015 = 2.25 np=150×0.015=2.25

STEP 4

Calculate n(1−p) n(1-p) n(1−p):n(1−p)=150×(1−0.015)=150×0.985=147.75 n(1-p) = 150 \times (1 - 0.015) = 150 \times 0.985 = 147.75 n(1−p)=150×(1−0.015)=150×0.985=147.75

STEP 5

Evaluate whether the normal distribution can be used:- np=2.25 np = 2.25 np=2.25 is less than 5. - n(1−p)=147.75 n(1-p) = 147.75 n(1−p)=147.75 is greater than 5.Since np<5 np < 5 np<5, the normal distribution cannot be used. The conclusion is:Do not use the normal distribution.

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QuestionDetermine whether the normal sampling distribution cain be used. The claim is $p<0.015$ and the sample size is $n=150$ . Use the normal distribution. Do not use the normal distribution.

1. The sample proportion is denoted by $\hat{p}$ .
2. The population proportion claim is $p < 0.015$ .
3. The sample size is $n = 150$ .
4. The normal approximation can be used if both $np$ and $n(1-p)$ are greater than or equal to 5.

1. Calculate $np$ and $n(1-p)$ .
2. Evaluate whether the normal distribution can be used based on the calculated values.

Calculate $np$ :
$np = 150 \times 0.015 = 2.25$

Calculate $n(1-p)$ :
$n(1-p) = 150 \times (1 - 0.015) = 150 \times 0.985 = 147.75$

Evaluate whether the normal distribution can be used:
- $np = 2.25$ is less than 5. - $n(1-p) = 147.75$ is greater than 5.
Since $np < 5$ , the normal distribution cannot be used.
The conclusion is:
Do not use the normal distribution.