Answer: The percentage of hotels in the city have a nightly cost of more than $200 is 21%
Step-by-step explanation:
Since the nightly cost of hotels in a certain city is normally distributed,
we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = the nightly cost of hotels.
µ = mean cost
σ = standard deviation
From the information given,
µ = $180.45
σ = $24.02
The probability that a hotel in the city has a nightly cost of more than $200 is expressed as
P(x > 200) = 1 - P(x ≤ 200)
For x = 200,
z = (200 - 180.45)/24.02 = 0.81
Looking at the normal distribution table, the probability corresponding to the z score is 0.79
Therefore,
P(x > 200) = 1 - 0.79 = 0.21
The percentage of hotels in the city have a nightly cost of more than $200 is
0.21 × 100 = 21%
The value of p is 13
5+11+6+4+7+10+p=43+p
43+p\7 =8
43+13=56
56/7=8
To find the state tax you have to take 7 percent of 124.95
Then add the tax to the amount he paid then add the charges for shipping
7/100=.07
124.95x.07=8.74
8.74+124.95+10=143.69
so he paid in total $143.69
Answer: 5/6
Step-by-step explanation: trust me on this
Answer:
0.0802
Step-by-step explanation:
Calculation for the p-value
Based on the information given this is a two tailed test
Hence,
p(z< -1.75)= 0.0401
p(z> 1.75) = 0.0401
Now let calculate the p-value
p-value=2* 0.0401
p-value= 0.0802
Therefore If the value of the test statistic z equals 1.75 the p-value will be 0.0802