An article claimed that only 38\88, percent of hotel visitors used the alarm clock provided. A hotel manager wanted to see whether that proportion applied to the hotel where she worked, so she took a random sample of 700700700 of the approximately 10{,}00010,00010, comma, 000 visitors to the hotel one month and checked whether they used their room alarm clock. Of the sampled visitors, only 35\55, percent used the alarm clock. Assuming that the article's 38\88, percent claim is also the true proportion for visitors to the hotel, what is the approximate probability that less than 35\55, percent of the sample used the alarm clock

Answer: P(p<0.35) = 0.05

Step-by-step explanation:

Khan told me

The approximate probability that less than 355 of the same used the alarm clock will be equal to 0.0485.

Probability is calculated as the proportion of favorable events to all potential scenarios of an event. The proportion of positive results, or x, for an experiment with 'n' outcomes can be expressed.

As per the given information given in the question,

P = 0.38

Q = 1 - P = 1 - 0.38 = 0.62

n = 700 and sample proportion, p = 0.35

Use the formula given below,

z = (p - P)/√(PQ/n)

z = (0.35 - 0.38)/√(0.38 × 0.62)/700

z = -0.03/0.018

z = -1.666

Now,

For P(X < 35%) = P(Z < -1.66)

= P(-∞ < Z < 0) - P(-1.66 < Z < 0)

= 0.5 - 0.4515 = 0.0485

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Your question seems incomplete, the complete question may be:

An article claimed that only 38% of hotel visitors used the alarm clock provided. A hotel manager wanted to see whether that proportion applied to the hotel where she worked, so she took a random sample of 700 and checked whether they used their room alarm clock. Of the sampled visitors, only 35% used the alarm clock.

What is the approximate probability that less than

35% of the sample used the alarm clock (rounded to the nearest hundreth)?

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