Question

# 1. For a standard normal distribution, find:   P(z > 2.32)   Keep four decimal places. 2. For...

1. For a standard normal distribution, find:

P(z > 2.32)

Keep four decimal places.

2. For a standard normal distribution, find:

P(-0.9 < z < 0.95)

3. For a standard normal distribution, given:

P(z < c) = 0.7622

Find c.

4. For a standard normal distribution, find:

P(z > c) = 0.1753

Find c

5. Assume that the readings at freezing on a batch of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find the probability of obtaining a reading less than -0.864°C.

P(Z<−0.864)=P(Z<-0.864)=

6. Assume that the readings at freezing on a batch of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find the probability of obtaining a reading between -0.495°C and 1.88°C.

P(−0.495<Z<1.88)=P(-0.495<Z<1.88)=

7. About % of the area under the curve of the standard normal distribution is outside the interval z=[−0.91,0.91]z=[-0.91,0.91] (or beyond 0.91 standard deviations of the mean).

8. Assume that the readings at freezing on a batch of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find P11, the 11-percentile. Round to 3 decimal places. This is the temperature reading separating the bottom 11% from the top 89%.

P11 = °C

9. z = -3 is what percentile?

percentile

State your answer to the nearest tenth of a percent. Don't include letters like "st", "nd", or "th" in your answer.

1:

The required probability, using z table, is

2;

The required probability, using z table, is

3:

Using z table we have

That is area left to z-score 0.71 is approximately 0.7622.

Hence, c = 0.71

4:

Using given equation we have

Using z table we have

That is area left to z-score 0.93 is approximately 0.8247.

Hence, c = 0.93

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