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For a certain population of sea turtles, 18 percent are longer than 6. 5 feet. A random sample of 90 sea turtles will be selected. What is the standard deviation of the sampling distribution of the sample proportion of sea turtles longer than 6. 5 feet for samples of size 90 ?.

Respuesta :

Based on the information given, it should be noted that the sampling distribution will be 0.0405.

Sampling distribution.

From the information given, it was stated that in a certain population of sea turtles, 18 percent are longer than 6. 5 feet and a random sample of 90 sea turtles will be selected.

Therefore, the following can be depicted:

  • p = 0.18
  • n = 90

Therefore, the standard deviation will be calculated thus:

= 0.18 × (1 - p) / 90.

= (0.18 × 0.82) / 90

= 0.00164

Therefore, the square root will be:

= ✓0.00164

= 0.0405.

In conclusion, the correct option is 0.0405.

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