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taurus [48]
3 years ago
11

Find the area to the right of the z-score 1.39 and to the left of the z-score 1.53 under the standard normal curve.1.2 1.3 1.4 0

.00 0.01 0.02 0.03 0.8849 0.8869 0.8888 0.8907 0.9032 0.9049 0.9066 0.9082 0.9192 0.9207 0.9222 0.9236 0.9332 0.9345 0.9357 0.9370 0.9452 0.9463 0.9474 0.9484 0.9554 0.9564 0.9573 0.9582 0.04 0.05 0.06 0.070.08 0.09 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 1.5 1.6 1.7 Use the value(s) from the table above.
Mathematics
1 answer:
svetoff [14.1K]3 years ago
6 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is  P( 1.39 <  z <  1.53 ) = 0.0193

Step-by-step explanation:

From the question we are told to obtain the area to the of the z-score 1.39 and to the left of the z-score 1.53, this is mathematically represented as

        P( 1.39 <  z <  1.53 ) =  P( z <  1.53 ) - P( z <  1.39 )

From the z table on the question the area under the normal curve to the left corresponding to 1.53 and  1.39  is

       P( z <  1.53 )  = 0.9370

and

      P( z <  1.39 )  = 0.9177

So

     P( 1.39 <  z <  1.53 ) = 0.9370 - 0.9177

=>  P( 1.39 <  z <  1.53 ) = 0.0193

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Suppose a certain capsule is manufactured so that the dosage of the active ingredient follows the distribution Y ~ N(μ = 10 mg,
lianna [129]

Answer:

(a) Probability that Y falls into the dangerous region is 0.0013.

(b) Probability that the mean Y-bar falls into the dangerous region is 0.00001.

Step-by-step explanation:

We are given that a certain capsule is manufactured so that the dosage of the active ingredient follows the distribution Y ~ N(μ = 10 mg, σ = 1 mg).

A dosage of 13 mg is considered dangerous.

Let Y = <u><em>dosage of the active ingredient </em></u>

The z-score probability distribution for normal distribution is given by;

                                 Z  =  \frac{ Y-\mu}{\sigma} } }  ~ N(0,1)

where, \mu = population mean = 10 mg

            \sigma = standard deviation = 1 mg

(a) Probability that Y falls into the dangerous region is given by = P(Y \geq 13 mg)

       P(Y \geq 13 mg) = P( \frac{ Y-\mu}{\sigma} } } \geq \frac{ 13-10}{1} } } ) = P(Z \geq 3) = 1 - P(Z < 3)  

                                                          = 1 - 0.9987 = <u>0.0013</u>

The above probability is calculated by looking at the value of x = 3 in the z table which has an area of 0.9987.

(b) We are given that a dosage of 13 mg is considered dangerous. And we sample 49 capsules at random.

Let \bar Y = sample mean dosage

The z-score probability distribution for sample mean is given by;

                                 Z  =  \frac{\bar Y-\mu}{\frac{\sigma}{\sqrt{n} } } } }  ~ N(0,1)

where, \mu = population mean = 10 mg

            \sigma = standard deviation = 1 mg

            n = sample of capsules = 49

So, Probability that the mean Y-bar falls into the dangerous region is given by = P(\bar Y \geq 13 mg)

          P(Y \geq 13 mg) = P( \frac{\bar Y-\mu}{\frac{\sigma}{\sqrt{n} } } } } \geq \frac{13-10}{\frac{1}{\sqrt{49} } } } } ) = P(Z \geq 21) = 1 - P(Z < 21)  

                                                             = <u>0.00001</u>

6 0
3 years ago
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